00 NQM11B SB TXT.prelim Q Maths 11B/ch.06.text.pdf · Solving quadratic equations 6.2 Algebraic solution of quadratics 6.3 Sketching quadratic functions 6.4 Finding simultaneous solutions

Contents

6.1

Solving quadratic equations

6.2

Algebraic solution of quadratics

6.3

Sketching quadratic functions

6.4

Finding simultaneous solutionsChapter summary

Chapter review

Syllabus subject matter

Introduction to functions

■

Concepts of function, domain and range

■

Ordered pairs, tables, graphs and equations asrepresentations of functions and relations

■

Graphs as a representation of the points whosecoordinates satisfy an equation

■

General shapes of functions, including: polynomials up todegree 4; reciprocal functions; absolute value functions

■

Practical applications: polynomials up to degree 2;reciprocal functions; absolute value functions

■

Solutions to simultaneous equations in two variables: graphically,using technology; algebraically (linear and quadratic equations only)

Quantitative concepts and skills

■

Calculation and estimation with and without instruments

■

Basic algebraic manipulations

■

Plotting points using Cartesian coordinates

■

Solutions of a quadratic equation

■

Graphs of quadratic functions

Syllabus Guide

Chapter 6

Parabolic functions

06 NQM11B SB TXT.fm Page 215 Sunday, May 17, 2009 7:57 PM

Parabolic functions

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New QMaths 11B

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6.1

Solving quadratic equations

You have examined linear functions in Chapter 3 of this textbook. Linear functions are one of a family of functions known as

polynomial functions

.

A linear function has the

standard (general) form

f

(

x

)

=

ax

+

b

and is a polynomial of degree 1. A linear equation has the standard form

ax

+

b

=

0. The graph of a linear function

f

(

x

)

=

ax

+

b

is a straight line and is equivalent to the graph of the corresponding linear equation

y

=

ax

+

b

.A

root

or solution of an equation is a value of the variable that makes the equation true. If a graph of a function is drawn, the root is the value of

x

where the graph crosses the

x

-axis, i.e.

f

(0). Because the roots of a function are the values for which

f

(

x

) = 0, they are also known as the

zeros

of the function. The zeros are also the

x

-intercepts of the function.

It is clear that a linear function can at most have one root.

A

quadratic function

is a polynomial of degree 2. Quadratic functions are also known as

quadratics

.

Quadratic functions and equations arise from the motion of projectiles near the Earth. When a stone is thrown, its path is parabolic. Determining where and when it will land involves the solution of quadratic equations. Quadratic equations arise from the solution of many area problems. The speeds of chemical reactions are quadratic functions, the resistance of a wire is a quadratic function of temperature and the energy of a moving object varies as a quadratic with speed. In this chapter, you will learn how to solve quadratic equations and the essential features of quadratic functions.

Polynomial functionsA polynomial function in x may be defined as:

f(x) = anxn + an − 1xn − 1 + an − 2xn − 2 + … + a1x1 + a0 For this definition, n is known as the degree of a polynomial and is the greatest power to which the variable (x) is raised.

!

x

Linear functions

Roots

y = f(x)

Standard form of a quadratic equation and functionA quadratic equation involves the second power of the variable (and no higher power).The standard (general) form of the quadratic equation is

ax2 + bx + c = 0 where a ≠ 0.A quadratic equation can be written in the form of a quadratic function as

f (x) = ax2 + bx + c

!


Chapter 6 Parabolic functions

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9780170136570

The shape of the graph of the quadratic in Example 1 is known as a

parabola

. The graphs of all quadratics have this distinctive shape, so quadratics are also known as

parabolic functions

.

A quadratic may have as many as two roots (zeros).

A distinctive feature of a parabola of the form

f

(

x

) =

ax

2

+

bx

+

c

is that it is symmetrical about a vertical axis passing through its

turning point

.

Use a graph to find the roots of the quadratic x2 − 3x − 5 = 0.

Solution

Construct a table of values for the function.

Now use the values to sketch the function, joining points with a smooth curve.The roots of the quadratic are f(0), i.e. the values of x where y = f(x) = 0.Using the graph, we can estimate that f (x) = 0 at about x = −1.2 and x = 4.2.

Write the equation. x2 − 3x − 5 = 0Express as a function. f(x) = x2 − 3x − 5We want to graph the function, so let y = f(x). y = x2 − 3x − 5

x −3 −2 −1 0 1 2 3 4 5 6

y 13 5 −1 −5 −7 −7 −5 −1 5 13

State the result. The roots of x2 − 3x − 5 = 0 are x ≈ –1.2 and x ≈ 4.2.

−4−1 4

y

12

8

4

−2−3 2 3 5 x

−8

y = x2 − 3x − 510

6

2

14

−6

1 6

Example 1

x

y = f(x)

No roots

Two rootsOne root

x

f(x)

Turningpoint

Turningpoint

Axis ofsymmetry


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218 New QMaths 11B9780170136570

When plotted on a Cartesian plane, values of f(x) are positive above the x-axis and negative below the x-axis.

All graphics calculators have a function. The following example shows how this function may be used to quickly find approximate values for the zeros of a function.

y

x

f(x) > 0

f(x) < 0

y = f(x)

TRACE

Use a graphics calculator to find approximate zeros of the functiony = f (x) = x2 − 7x + 9

SolutionTo find the zeros we can use the fact that f (x) changes sign from positive to negative, or negative to positive, when it crosses the x-axis.Enter the function Y = X2 − 7X + 9 in Y1.

Set the (or V-Window) to minimum and maximum domain values of 0 and 6 and range values of −4 and 4.

the function. Looking at the graph, you can see that there are zeros near 2 and 5.

Select the function and use the and keys to move the cursor close to the zero (root) near 2.

When you get very close to the zero, you will notice that the value of Y changes from positive to negative (or vice versa depending on the direction in which you are moving).

The screen dumps show how Y changes from 0.111…

to −0.061… with a single press of the key. (These

values may vary slightly depending on the calculator used.)This means that the zero occurs at a value of X somewherebetween 1.666 … and 1.714 … If required, a more exact

value can be found by zooming in and again using the function.

Note: When using the function in this example, the cursor of the calculatorappears to ‘jump’ along the curve in small increments or steps. This is because most calculators use discrete (but small) increments to calculate the points on a curve. We know that the function is continuous, but the calculator draws it using discrete increments. Because the increments are small, it doesn’t really matter for most situations.

The same procedure may now be used to locate the other zero at X ≈ 5.3.

The zeros of f (x) = x2 − 7x + 9 are about 1.7 and 5.3.

WINDOW

GRAPH

TRACE

TRACE

TRACE

Example 2


Chapter 6 Parabolic functions 2199780170136570

Most graphics calculators have inbuilt functions that can be used to find the roots of equations.

Use a graphics calculator to draw a graph and hence find approximate solutions for3x2 + x = 8.

SolutionExpress in function form: f(x) = 3x2 + x − 8. Enter the function y = 3x2 + x − 8 in Y1.

Set the (or V-Window) to the default minimum and maximum domain and range values of −10 and 10.

Casio fx-9860G AUSelect GRAPH from the main menu and select DRAW to graph the function.As can be seen from the display, the zeros are near −2 and 1.5.

Change the V-Window so that Xmin = −2 and Xmax = 2.GRAPH the function again.

Select G-Solv and then select ROOT to find the zeros of the function more exactly.

Texas Instruments TI-84

the function. As can be seen from the display, the zeros are near −2 and 1.5.

Change the so that Xmin = −2 and Xmax = 2.

the function again.

Select 2: zero from the CALC menu to find the zerosmore exactly.Type −2 for the left bound and −1 for the right bound, then

for the guess. This gives a value of X ≈ −1.8081.

Sharp EL-9900See the instructions given on the CD-ROM.

This gives a value of X ≈ −1.8081. Similarly, the other zero is X ≈ 1.4748.

The solutions of 3x2 + x − 8 = 0 are about −1.81 and 1.47.

Repeating the procedure, with left and right bounds of say 1 and 2 respectively, gives the other zero as X ≈ 1.4748.

The solutions of 3x2 + x − 8 = 0 are about −1.81 and 1.47.

WINDOW

GRAPH

WINDOW

GRAPH

ENTER

Example 3

Calculator Instructions


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220 New QMaths 11B9780170136570

You can also find solutions to some quadratic equations by systematic trial and error. This is the simplest numerical method for solution. It involves the process of systematically trying different values of x until a solution is found.

In some cases, systematic trial and error will not give an exact solution. If so, we must decide how accurate we want the solution. This can be a tedious process ‘by hand’. Using a graphics calculator, a spreadsheet or a computer algebra system helps to make the calculations easier.

Find the solutions to 5x2 − x − 48 = 0 by systematic trial and error.

SolutionCalculate 5x2 − x − 48 for a range of values of x. Look for x values so that 5x2 − x − 48 = 0.

We know the general shape of the graph.Make a rough sketch.Because the value of 5x2 − x − 48 changes from negative to positive between x = 3 and x = 4, there must be another solution between 3 and 4.

x −4 −3 −2 −1 0 1 2 3 4

5x2 − x − 48 36 0 −26 −42 −48 −44 −30 −6 28

One solution is immediately clear. Write it down. x = −3

Write down the function. f(x) = 5x2 − x − 48Substitute an x value between 3 and 4, say 3.4. f(3.4) = 5 × (3.4)2 − 3.4 − 48

= 6.43.4 was too high. Looking at the graph, try a smaller value, say 3.2.

f(3.2) = 5 × (3.2)2 − 3.2 − 48 = 0

This is the second solution. x = 3.2Write the answer. The solutions are x = −3 and x = 3.2.

f (x)40

20

−3−4 1 2 3 x

−30

−50

f(x) = 5x2 − x − 4830

10

−40

4−2

−20−10

Example 4

Use a spreadsheet to help find solutions to 3x2 = 6x + 12 correct to 2 decimal places.

SolutionChange the equation so that it is in the standard form: 3x2 − 6x − 12 = 0.In the spreadsheet, type ‘Start’ into cell A1, ‘Increment’ into A2, −4 into B1, 1 into B2,‘x’ into A3 and ‘Value’ into B3, then type in the following formulas:Type the formula =B1 into cell A4.Type the formula =A4+$B$2 into cell A5.Type the formula =3*A4^2-6*A4-12 into cell B4 to enter the function.

Example 5



Your spreadsheet should look like this:

Now copy cell A5 to cells A6:A14 and B4 to cells B5:B14.

One solution must lie between x values −2 and −1 and the other between 3 and 4.

First rootTo find the solution between −2 and −1 more exactly, type −2 into B1 and 0.1 into B2. Then part of your spreadsheet should look like this:

The solution must lie between −1.3 and −1.2. Type −1.3 into B1 and 0.01 into B2.

The solution must be between −1.24 and −1.23. Type −1.24 into B1 and 0.001 into B2.

The solution is closer to −1.24 than −1.23.


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222 New QMaths 11B9780170136570

Exercise 6.1 Solving quadratic equations 1 In each of the following, use the graph to estimate the roots of the function correct to

1 decimal place.a b

c d

2 Plot graphs of the following quadratics to find the approximate roots of each.a f (x) = x2 + x − 12 b y = 5 − 4x − x2 c y = x2 − 4x − 6

d f(x) = 3x + 7 − 2x2 e f (x) = 2x2 + 3x − 8 f y = − 4x − 7

3 Use a graphics calculator to draw a graph of each of the following and hence find the approximate roots.

a f (x) = 2x2 − 5x − 10 b f (x) = 8 − − 4x c y = 5x2 − 14x − 11

d y = 6x + 31 − 3x2 e y = 4x2 − 8x − 26 f f(x) = + 7x − 15

Second rootSimilarly, work on the other root, between 3 and 4. You should obtain:

The solution is closer to 3.24 than 3.23. The solutions to 3x2 = 6x + 12, correct to 2 decimal places, are −1.24 and 3.24.

Additional Exercise

6.1

2−3 1−1−2 x

10

−30−25−20

5

−5−10

y15

−15

42−2 x

5

−20

−5−10

y10

−15

6

−8−6

42

−2

y8

−4

6−4 42−2 x

−4 42−2 x

10

2

864

y12

2

x2

2-----

x2

2-----

x2

5-----



4 A spreadsheet is being used to find the approximate roots of quadratic functions in a similar process to that used in Example 5. For each of the following spreadsheet screen dumps, state two values between which the root lies.a b

c d

5 Find the zeros, correct to 2 decimal places, of each of the following quadratics using systematic trial and error.a f (x) = 3x2 − 8x − 3 b f (x) = 3x2 − 2x − 15 c y = 11 + 3x − 6x2 d y = 13 − 2x − 2x2 e f (x) = x2 + 5x − 7 f y = x2 − 2x − 10

6 Use a spreadsheet and systematic trial and error to find the roots of each of the following quadratics correct to 2 decimal places.

a f (x) = 2x2 − 9x − 13 b y = 4x − 3x2 + 17 c y = + 3x − 1

d y = + 3x + 15 e f (x) = 9x2 − 5x − 17 f f (x) = 17 − 5x − 3x2

6.2 Algebraic solution of quadraticsYou saw in the previous section that solving quadratic equations using graphical and trial-and-error methods can be tedious and may not provide exact answers. We can also solve quadratic equations algebraically. The first step in this process requires the quadratic expression to be factorised.

Before we see how to factorise quadratic expressions, it is helpful to review the process for multiplying binomial expressions.

x2

3-----

x2–4

--------


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224 New QMaths 11B9780170136570

Factorising quadratics is a matter of reversing the product of the binomials that result in a quadratic expansion as shown in Example 6. Simple quadratics in the form ax2 + bx + c may be factorised using either the decomposition method or the cross method.

The decomposition method is the reverse of the distributive law method of quadratic expansion. We ‘decompose’ the middle term into two terms.■ The coefficients of the two new terms must add to give the coefficient of the quadratic’s

middle term.■ Their product must be the same as the product of the coefficients of the first and last terms.■ The middle term is decomposed and the factorisation completed using the grouping method.

The cross method is the reverse of the FOIL method of quadratic expansion.■ List factors of the x2 term underneath each other on the left.■ List the factors of the last term under each other on the right.

Expand the following products.a (x + 4)(2x + 7) b (3x − 5)(4x + 9)

Solution

Method 2 (FOIL)Using the FOIL method, we multiply the First, Outer, Inner and Last terms of the product as shown.First terms are x and 2x.Outer terms are 4 and 7.Inner terms are 4 and 2x.Last terms are 4 and 7.

a Method 1 (distributive law)Write the product. Apply the distributive law a(b + c) = ab + ac.

(x + 4)(2x + 7) = x(2x + 7) + 4(2x + 7)

Apply the distributive law again. = 2x2 + 7x + 8x + 28Simplify. = 2x2 + 15x + 28

Now multiply the terms. (x + 4)(2x + 7) = x × 2x + x × 7 + 4 × 2x + 4 × 7= 2x2 + 7x + 8x + 28

Simplify. = 2x2 + 15x + 28

b Method 1 (distributive law)Write the product. Apply the distributive law a(b + c) = ab + ac.

(3x − 5)(4x + 9) = 3x(4x + 9) − 5(4x + 9)

Apply the distributive law again. = 12x2 + 27x − 20x − 45Simplify. = 12x2 + 7x − 45

Method 2 (FOIL)Multiply the terms. (3x − 5)(4x + 9) = 3x × 4x + 3x × 9 − 5 × 4x − 5 × 9

= 12x2 + 27x − 20x − 45Simplify. = 12x2 + 7x − 45

First Outer

Inner Last

(x + 4)(2x + 7)

Example 6



■ Use cross products to obtain the outer and inner products.■ Add the outer and inner products to check whether the factors are correctly listed to produce

the middle term of the quadratic.■ If the middle term is incorrect, rearrange the factors or try new factors until the correct middle

term is obtained.■ The factors on the left will be the coefficients of the first terms in the binomial factors of the

quadratic. The factors on the right will be the coefficients of the last terms in the binomial factors.

■ Form the binomial factors.

It doesn’t matter which method you choose. However, it is probably best to use only one of these methods, as using both tends to be confusing. Your teacher may have a preference for one of the methods. Some people choose the method according to the coefficient of the first term: they use the cross method if it is 1, and the decomposition method for other cases.

Factorise these quadratics.a x2 − 11x + 24 b 20x2 − 31x − 9

Solutiona Decomposition method

Consider the quadratic x2 − 11x + 24. The coefficient of the middle term is −11.The product of the coefficients of the first and last terms is +24.We want two numbers that add to −11 and multiply to +24. Since the product is positive, the numbers must have the same sign. Since they add to give −11, they must both be negative.Find numbers whose product is 24. 24 = 24 × 1 = 12 × 2 = 3 × 8 = 6 × 4Select the pair that add to −11. −3 + −8 = −11 Now write the expression. x2 − 11x + 24Decompose −11x using −3 + −8 = −11. = x2 − 3x + −8x + 24Factorise. Be careful with signs! = x(x − 3) − 8(x − 3)Factorise again: common factor is (x − 3). = (x − 8)(x − 3)

Cross methodConsider the quadratic x2 − 11x + 24.Possible factors of 1 are 1 × 1.Possible factors of 24 are 24 × 1, −24 × −1, 8 × 3, and so on.

Set up the ‘cross’ with 1 × 1 on the left, and try the factors of the second term. on the right.

The coefficients of the first terms are 1 and 1.The coefficients of the last terms are −8 + −3.

Now write the result. x2 – 11x + 24 = (x – 8)(x − 3)

1

1

−6

−4

−10−8

−3

−11 That’s it!

1

1

−6

−4

−10−8

−3

−11(1x + −8)

(1x + −3)

Example 7


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226 New QMaths 11B9780170136570

Once a quadratic expression is factorised, it can be solved by using the null factor law.

b Decomposition methodConsider the quadratic 20x2 − 31x − 9. The coefficient of the middle term is −31.The product of the coefficients of the first and last terms is −180.We want two numbers that add to −31 and multiply to −180. Since the product is negative, the numbers must have opposite signs: one positive and one negative.

Cross methodConsider the quadratic 20x2 − 31x − 9.Possible factors of 20 are 20 × 1, 10 × 2 and 5 × 4.Possible factors of −9 are 9 × −1, −9 × 1 and 3 × −3.

List products. 180 = 10 × 18 = 9 × 20 = 3 × 60 = 5 × 36 = 2 × 90 = 1 × 180

Select the pair that add to −31. 5 + −36 = −31 Write the expression. 20x2 − 31x − 9Decompose −31x using 5 + −36 = −31. = 20x2 + 5x + −36x − 9Factorise. = 5x(4x + 1) − 9(4x − 1)Factorise again: common factor is (4x − 1). = (5x − 9)(4x + 1)

Set up a ‘cross’ with 10 × 2 on the left,and try factors of the second term onthe right.

Try a new ‘cross’ with 5 × 4 on the left.The binomials are (5x + −9) and (4x + 1).

Write the result. 20x2 − 31x − 9 = (5x − 9)(4x + 1)

10

2

9

−1

8−1

9

88−9

1

−81

−9

−883

−3

−24−3

3

24 Nothing works!

5

4

9

−1

31−9

1

−31 That’s it!

Null factor lawIf ab = 0, then either a = 0, b = 0 or a = b = 0.

!

Solve the following quadratic equations.a (x − 3)(x + 4) = 0 b 5p2 + 2p − 70 = 2p2 + 14p − 7

Solutiona Write down the equation. (x − 3)(x + 4) = 0

Apply the null factor law. (x − 3) = 0 or (x + 4) = 0Solve the resulting equations. x = 3 or x = −4

Example 8



Completion of the square is an alternative algebraic method of solving quadratic equations. Completion of the square is a useful technique because it also works in many cases where factorisation fails. This method relies on the factorisation x2 + 2ax + a2 = (x + a)2 to make a perfect square of the variable parts of the equation.

The steps are as follows.

b Write down the equation. 5p2 + 2p − 70 = 2p2 + 14p − 7Express in standard form. 3p2 − 12p − 63 = 0Take out the common factor. 3(p2 − 4p − 21) = 0Factorise the quadratic. 3( p − 7)( p + 3) = 0Solve using the null factor law. p = 7 or p = −3

Solve x + 5 = − .

SolutionWrite down the equation. x + 5 = −

Multiply by the denominator to clear the fraction. x2 + 5x = −6Express in standard form. x2 + 5x + 6 = 0Factorise. (x + 2)(x + 3) = 0Solve using the null factor law. x = −2 or x = −3

6x---

6x---

Example 9

Completion of the square1 Start with the standard form of the quadratic equation.

2 If necessary, divide to make the coefficient of x2 equal to 1.

3 Put the constant term on the right-hand side.

4 Find a (from the pattern x2 + 2ax + a2 = (x + a)2) by halving the linear coefficient, i.e. the coefficient of x.

5 Add a2 to both sides.

6 Write the left-hand side as a perfect square.

7 Take the square root, writing two equations to show the positive and negative cases.

8 Find the final solutions.

!


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228 New QMaths 11B9780170136570

Completion of the square is a technique that can be applied to any quadratic equation. This means we can derive a formula for the general quadratic equation using this method. The general derivation is shown on the CD-ROM.

Solve x2 + 6x − 16 = 0 by completing the square.

SolutionWrite down the equation. x2 + 6x − 16 = 0Move the constant term to the RHS. x2 + 6x = 16The coefficient of x is 6, so a = 3.Add a2 to both sides. x2 + 6x + 9 = 16 + 9Use the pattern x2 + 2ax + a2 = (x + a)2 to writethe LHS as a perfect square.

(x + 3)2 = 25

Take the square root of both sides. x + 3 = 5 or x + 3 = −5Solve the resulting equations. x = 2 or x = −8

Example 10

Solve 3p2 = 12 − 15p correct to 2 decimal places by completing the square.

SolutionWrite down the equation. 3p2 = 12 − 15pExpress with constant term only on the RHS. 3p2 + 15p = 12Divide throughout by the common factor 3. p2 + 5p = 4

Now a = 5 � 2 = , so a2 = .

Add a2 to both sides. p2 + 5p + = 4 +

Use x2 + 2ax + a2 = (x + a)2 to factorise the LHS, and simplify the RHS. =

Take the square roots. p + = ±

Solve the resulting equations. p = or

Evaluate correct to 2 decimal places. ≈ 0.70 or −5.70Write the solutions. p ≈ 0.70 or p ≈ −5.70

52--- 25

4------

254

------ 254------

p 52---+⎝ ⎠

⎛ ⎞ 2 414------

52--- 41

2----------

41 5–2

------------------- − 41 5–2

-----------------------

Example 11

Extra Material

The quadraticformula

Quadratic formulaFor any quadratic equation ax2 + bx + c = 0 where a ≠ 0, the solutions are given by

x = −b b2 4ac–±2a

---------------------------------------

!



For f(x) = ax2 + bx + c, the expression b2 − 4ac, called the discriminant, is a useful tool for determining the number of roots a quadratic equation has, and whether the roots (if any) are rational or irrational numbers. The discriminant is explored on the CD-ROM.

Exercise 6.2 Algebraic solution of quadratics1 Factorise these quadratics.

a 3x2 − 11x + 10 b x2 + 6x − 16 c 7x2 + x − 6d 2x2 − 11x + 12 e 4x2 − 13x − 12 f 8x2 + 18x + 9 g 5x2 + 36x + 7 h 3x2 − 14x − 49 i 7x2 − 52x − 32j 5x2 − 44x + 32 k 20x2 − 37x − 18 l 25x2 − 30x − 16

2 Solve the following by factorisation where necessary.a 2x(x − 12) = 0 b (x + 2)(x − 5) = 0 c x2 − 8x = 0d 4x2 + 6x = 0 e x2 + 9x + 14 = 0 f x2 − 8x + 15 = 0

Solve 3x2 = 4x + 12 using the quadratic formula.

SolutionWrite the equation. 3x2 = 4x + 12Express in standard form. 3x2 − 4x − 12 = 0Compare with ax2 + bx + c = 0. a = 3, b = −4, c = −12

Write the quadratic formula. x =

Substitute for a, b and c. =

Evaluate. =

Simplify. =

Simplify the surd. =

=

Factorise. =

Simplify. =

Write as two solutions. = or

If approximate solutions are needed, evaluate. ≈ 2.77 or −1.44Write the approximate solutions. x ≈ 2.77 or x ≈ −1.44

−b b2 4ac–±2a

---------------------------------------

4– 4–( )2 4– 3× 12–×±–2 3×

----------------------------------------------------------------------

4 16 144+±6

-----------------------------------

4 160±6

----------------------

4 16 10×±6

------------------------------------

4 4 10±6

-------------------------

4 1 10±( )6

---------------------------

2 1 10±( )3

---------------------------

2 1 10+( )3

---------------------------- 2 1 10–( )3

----------------------------

Example 12

Extra Material

The discriminant

Additional Exercise6.2


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230 New QMaths 11B9780170136570

3 Solve by factorisation.a 3x2 + 5x − 12 = 0 b 2x2 − 24x + 22 = 0 c 3x2 + 15x + 12 = 0d 4x2 − 28x − 32 = 0 e 2x2 + 14x + 20 = 0 f x2 − 2x = 3

4 Solve by factorisation.

a x − 4 = b x2 − x = 6 c x2 − x = 12

d x2 − 6x = −9 e 2x2 + 5x + 2 = 0 f 3x2 + 10x + 3 = 0

5 Solve by factorisation.a 5a2 + 8a − 4 = 0 b 2v2 − 19v + 42 = 0 c n2 − 64 = 0d 4c2 − 28c + 49 = 0 e 14e2 + 33e + 18 = 0 f 9p2 + 10p − 16 = 0

6 Solve by factorisation.a 2b2 = 11b − 14 b 10d2 + 10d + 10 = 3d2 − 9d

c 6x2 − 5x − 50 = 5x2 − 5x − 1 d 5a2 = 4(a2 + 2a − 4)e 4g(g + 6) = g − 15 f 16(x2 − 1) = 14x − 19

7 Solve by factorisation.

a x2 − 2x + 2 = b + 5 = − 2x

c + 2x = 1 − d + = +

e 16x = f 4x − = x + 2

8 Solve by completing the square. Express irrational answers correct to 2 decimal places.a k2 + 4k − 5 = 0 b v2 + 2v − 35 = 0c j2 = 9j − 8 d x2 = x + 132

e m2 = − f b2 + 4b − 6 = 0

9 Solve the following equations using the quadratic formula. Where appropriate, give both exact and approximate answers correct to 2 decimal places.a x2 + 14x + 45 = 0 b x2 + 20x + 51 = 0c x2 = 7x + 294 d x2 + 4x = 7e x2 − 6x = 21 f x2 = 13x + 140

10 Solve the following using the quadratic formula, correct to 2 decimal places.a 3f 2 + f − 14 = 0 b 5t2 = 9t + 18c 6x2 = 5x + 1 d x2 − x − 5 = 0e 9p2 = 15p + 5 f 2d2 + 3d = 6

11 Solve the following equations for x, using the most appropriate method.a x2 + 6x − 10 = 0 b 18x2 + 15x = 12c x2 − 4x − 13 = 0 d x2 − 6.146x + 9.328 = 0

e 1 − + = 0 f − =

g x2 + 29x + 6 = 0 h 3x2 − 5.6x + 0.56 = 0

5x---

1 3x2+4

------------------ x2 9x+4

------------------ 4x 1+2

----------------

3x2

2-------- 1

3---

x3--- x2

4----- x

3--- x2

6----- x 2+

4------------

2x 54–x 4–

------------------ 48x 3–------------

89--- 2m

3-------

x2--- x2

24------ 1

x 1+------------ 4

x--- 1

x 3+------------



Modelling and problem solving12 The length of a rectangular lawn is 0.9 m more than its width, and its area is 16.2 m2. Write

an equation and solve it to find the dimensions of the lawn.

13 A square room is 3 m high and 10 L of paint is needed to paint all the walls and ceiling. If 1 L of the paint will cover 16 m2, write an equation and find the length of the room.

14 The length of a closed box is three times its height, and the box is 4 cm wide. If the total surface area is 88 cm2, find the dimensions of the box.

15 Marnie can walk 1 km/h faster than Jon. She completes a 20 km hike 1 hour before him. Write an equation and solve it to find their walking speeds.

16 A bath toy bought for $x is sold for $24, making x% profit. What was the cost of the toy?

17 A tank can be filled in 6 hours using two pipes. The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone. How long would each pipe alone take?

18 A cyclist travelled at a certain speed to a town 30 km away, then reduced speed by 4 km/h and came back. If the cyclist had travelled at a steady 8 km/h for the whole trip, the total time would have been half an hour less. What was the initial speed of the cyclist?

19 The total resistance R of two resistors R1 and R2 in parallel is given by

= +

A particular resistor is placed in parallel with one 20 ohms greater in resistance. The total resistance is 24 ohms. What are the resistances of the two resistors?

16.2 m2

1R--- 1

R1------ 1

R2------


Parabolic functions

232 New QMaths 11B9780170136570

6.3 Sketching quadratic functionsIn this chapter you have already seen that the graph of a quadratic functions has a distinctive parabolic shape, which is symmetrical about a vertical axis drawn through the turning point. The turning point is so named because it is the point on the curve where the slope of the graph changes from positive to negative or from negative to positive.

The turning point of a quadratic function is either a maximum or a minimum as shown in the diagram above.

The graphics calculator is well suited to allow you to see how the shape of the graph of a quadratic function can change. We will consider the quadratic function y = f(x) = a(x − p)2 + q.It is important to have the function expressed in this form rather than the general form because it is easier to see any relationships that emerge.

Horizontal shiftWe begin by concentrating on the behaviour of the function y = a(x − p)2 + q as the value of p changes, so let’s fix a to be 1 and q to be 0. This means we will look at the function y = (x − p)2.Enter the function Y = X2 in Y1. (This is Y = (X − 0)2.)

Set the (or V-Window) to minimum and maximum domain values of −6 and 6 and range values of 0 and 10.

the function.

Now, on the same set of axes, Y = (X + 3)2 and Y = (X − 3)2.Watch the graphs carefully as they are drawn.Repeat this procedure with Y = X2, Y = (X + 5)2 and Y = (X − 5)2.Comparing the graphs you have drawn with y = (x − p)2, you can see that the value of p will slide or translate the graph of y = x2 horizontally p units to the right or left: to the right if p is positive, to the left if p is negative.You can also see that, for the function expressed in the form y = (x − p)2, the value of p is thex-coordinate of the turning point.

Vertical shiftThis time, we concentrate on the behaviour of the function y = a(x − p)2 + q as the value of q changes, so let’s fix a to be 1 and p to be any number, say 2. This means we will look at the function y = (x − 2)2 + q.Clear all graphs and enter the function Y = (X − 2)2 in Y1. (This means q = 0.)

Negative slope

y

x

Minimum

Maximum

Positive slopeNegative slope

Positive slope

Axi

s

Axi

s

Technology

p = 3p = −3

WINDOW

GRAPH

GRAPH



Set the (or V-Window) to minimum and maximum domain values of −3 and 5 and range values of −6 and 10.

the function.Now, on the same set of axes, Y = (X − 2)2 + 5 and Y = (X + 2)2 − 5.Watch the graphs carefully as they are drawn.Repeat this procedure with values of q other than 5.Comparing the graphs you have drawn with y = (x − 2)2 + q, you can see that the value of q will slide or translate the graph of y = x2 vertically q units up or down.You can also see that, for the function expressed in the form y = (x − p)2 + q, the value of q isthe y-coordinate of the turning point.

DilationThis time, we concentrate on the behaviour of the function y = a(x − p)2 + q as the value of a changes, so let’s fix p and q to be any numbers, say 2 and 5 respectively. This means we will look at the function y = a(x − 2)2 + 5.Clear all graphs and enter the function Y = (X − 2)2 + 5 in Y1. (This means a = 1.)

Set the (or V-Window) to minimum and maximum domain values of −2 and 7 and range values of 0 and 15.

the function Y = (X − 2)2 + 5.

Now, on the same set of axes, Y = 3(X − 2)2 + 5, Y = 6(X − 2)2 + 5 and Y = 0.5(X − 2)2 + 5.

Watch the graphs carefully as they are drawn.Repeat this procedure with other positive values of a.Comparing the graphs you have drawn with y = a(x − 2)2 + 5, you can see that the value of a dilates the graph. The value of a has no impact on the turning point.■ If a � 1, the graph of y = a(x − 2)2 + 5 is narrower than the graph of y = (x − 2)2 + 5 (has

an enlargement or dilation factor �1).■ If a � 1 (and a � 0), the graph of y = a(x − 2)2 + 5 is wider than the graph of

y = (x − 2)2 + 5 (has an enlargement or dilation factor � 1).

A value of a � 0 will cause the parabola to be inverted. Try a similar investigation with negative values of a. Other than the inversion of the parabola, you will obtain similar results.

The turning point of a quadratic function can be found without graphing by using a variation of the completing the square method. This method was used for solving quadratic equations in section 6.2. Finding the turning point of a quadratic function y = ax2 + bx + c follows similar steps, except that the constant term remains on the same side as the x2 and x terms.

WINDOW

GRAPH q = 5

q = −5

GRAPH

WINDOW

GRAPHa = 6 a = 3 a = 0.5a = 1

GRAPH

Shape of quadratic graphsy = a(x − p)2 + q

Dilation Horizontal shift Vertical shifta � 1, narrows parabola x-coordinate of y-coordinate of a � 1, expands parabola turning point turning point

!


Parabolic functions

234 New QMaths 11B9780170136570

We can also find the coordinates of the turning point by using the fact that the graph of a quadratic is symmetrical about an axis passing through the turning point. Consider the graph of f (x) = ax2 + bx + c as shown on the right.The turning point (xt, f (xt)) lies on the axis xt = k.The graph has zeros at x1 and x2.

Using the quadratic formula, the zeros are

x =

Using symmetry k = (x1 + x2)

=

= =

But k = xt

Finding the turning point using completion of the squareFor the quadratic function y = ax2 + bx + c:

1 Factorise the RHS so that the coefficient of x2 is 1.

2 Complete the perfect square by adding the square of half the coefficient of the x term, and subtracting the added term.

3 Separate out the perfect square and then factorise it.

The process changes y = ax2 + bx + c to y = a(x − p)2 + q to give the turning point (p, q).

Maximum and minimum points■ For positive coefficients of x2, the turning point is a minimum.■ For negative coefficients of x2, the turning point is a maximum.

!

Alternative Method

Find the turning point of f(x) = 8x + 4 − 2x2 and state whether it is a maximum or a minimum.

SolutionWrite the function. f (x) = 8x + 4 − 2x2

Rearrange and factorise to make the coefficient of the x2 term equal to 1.

= −2(x2 − 4x) + 4

Add and subtract the square of half the coefficient of the x term.

= −2(x2 − 4x + 4 − 4) + 4

Separate out the perfect square. = −2(x2 − 4x + 4) + −2 × −4 + 4Factorise the perfect square and simplify. = −2(x − 2)2 + 12Write the turning point. Turning point is (2, 12).Coefficient of x2 � 0. Turning point is a maximum.

Example 13

x

y

x1 x2

ZeroZero

Turning point(xt, f (xt))

xt = k

f (x) = ax2 + bx + c

b b2 4ac–±–2a

-------------------------------------

12---

12--- −b b2 4ac–+

2a--------------------------------------- −b b2 4ac––

2a---------------------------------------+⎝

⎛⎠⎞

12--- 2b–

2a---------⎝ ⎠

⎛ ⎞ b–2a------



So xt =

∴ Turning point (xt, f(xt)) =

The following summary of the various features of quadratic graphs that will be useful when you draw the graphs of quadratic functions.

b–2a------

b–2a------ f, b–

2a------⎝ ⎠

⎛ ⎞⎝ ⎠⎛ ⎞

Turning pointThe turning point of the quadratic f(x) = ax2 + bx + c where a ≠ 0 is

b–2a------ f, b–

2a------⎝ ⎠

⎛ ⎞⎝ ⎠⎛ ⎞

!

Identify the turning point of f (x) = x2 + 4x − 21 and state its type.

SolutionWrite the function. f(x) = x2 + 4x − 21Compare with ax2 + bx + c. a = 1, b = 4, c = −21

Calculate . = = −2

Calculate f . f = (−2)2 + 4 × −2 − 21 = −25

Write the turning point. Turning point is (−2, −25).Coefficient of x2 � 0. Turning point is a minimum.

b–2a------ b–

2a------ 4–

2------

b–2a------⎝ ⎠

⎛ ⎞ b–2a------⎝ ⎠

⎛ ⎞

Example 14

Quadratic graphsThe graph of a quadratic function is a parabola.The graph has the following shape:■ If the coefficient of the x2 term is positive, the

function decreases to a minimum value at theturning point, and then increases.

■ If the coefficient of the x2 term is negative, thefunction increases to a maximum value at theturning point, and then decreases.

The graph is symmetrical about a vertical line calledthe axis passing through the turning point.If the function has zeros, these are symmetrical about the turning point.For graphs of quadratic functions, the domain (possible x values) is the set of real numbers.The range (possible f (x) or y values) must be greater than or equal to the minimum value, or less than or equal to the maximum value.

Zero

Minimum Maximum

Axi

s

Axi

s

Zero x

y

!


Parabolic functions

236 New QMaths 11B9780170136570

The essential features of the graph of a quadratic are the turning point, the zeros (if they exist), the axis of symmetry and the y-intercept.■ The turning point of a quadratic function may be found by completing the square.■ The zeros (if they exist) may be found by factorisation, completing the square or using the

quadratic formula.■ The axis of symmetry is the vertical line that passes through the turning point.■ The y-intercept is the value when x = 0.

The graph of a quadratic function can be sketched using this information. If a domain is stated, then the endpoints and range also should be found.

Sketch the function y = x2 − 4x − 12 and state the range.

Solution

Sketch the graph.

Turning pointWrite the function. y = x2 − 4x − 12Complete the square. = x2 − 4x + 4 − 4 − 12Factorise the perfect square and simplify. = (x − 2)2 − 16State the coordinates of the turning point. Turning point is (2, −16).Coefficient of x2 term � 0. Turning point is a minimum.

ZerosLet the function be zero. x2 − 4x − 12 = 0Factorise. (x − 6)(x + 2) = 0Solve. x = 6 or x = −2State the result. x-intercepts are −2 and 6.

Other featuresWrite the function. y = x2 − 4x − 12Let x = 0. y = −12State the result. y-intercept is −12.Axis of symmetry passes through the turning point.

Axis of symmetry is x = 2.

No domain is stated, so the implied domain is the real numbers.

The range is y � −16.

y

6 x

(2, −16)

−2

−12y = x2 − 4x − 12

Example 15



Sketch the function f (x) = 4x − 3x2 + 5 for −1 � x � 3 and state the range.

Solution

Sketch the graph.

Turning pointWrite the function. f (x) = 4x − 3x2 + 5

Rearrange and factorise. = −3(x2 − x) + 5

Complete the square. = −3(x2 − x + − ) + 5

Separate out the perfect square. = −3(x2 − x + ) + −3 × − + 5

Factorise the perfect square and simplify. = −3(x − )2 + 6

State the result. Turning point is ( , 6 ).

Coefficient of x2 term � 0. Turning point is a maximum.Zeros

Let the function equal zero. 4x − 3x2 + 5 = 0Express in standard form. −3x2 + 4x + 5 = 0Compare with ax2 + bx + c = 0. a = −3, b = 4, c = 5

Use the quadratic formula. x =

Substitute for a, b and c. =

Simplify. =

State the exact solutions. =

Approximate solutions are useful for graphing. ≈ −0.786 or 2.120State the result. x-intercepts are −0.786 and 2.120.

Other featuresWrite the function. f (x) = 4x − 3x2 + 5Find f(0). f (0) = 5State the result. y-intercept is 5.Axis of symmetry passes through the turning point. Axis of symmetry is x = .Investigate f(x) at extreme domain values. f (−1) = −2 f (3) = −10State the result. Endpoints are (−1, −2) and (3, −10).

State the result. The range is −10 � y � 6 .

43---

43--- 4

9--- 4

9---

43--- 4

9--- 4

9---

23--- 1

3---

23--- 1

3---

b b2 4ac–±–2a

-------------------------------------

4– 42 4– 3–× 5×±2 3–×

------------------------------------------------------

4 76±–6–

----------------------

2 19±3

------------------

23---

( , 6 )

x

y

5

2.12

(−1, −2)

−0.79

(3, −10)

23

13

y = 4x − 3x2 + 5

13---

Example 16


Parabolic functions

238 New QMaths 11B9780170136570

We can use the graphics calculator to draw graphs of quadratic functions and to find the key points.

Draw the graph of y = x2 − 5x + 4 on a graphics calculator and find the key points.

SolutionFor all calculatorsEnter the function y = x2 − 5x + 4 in Y1. Set the (or V-Window).

Draw the graph with a domain from −1 to 6 and a range from −3 to 3.

Casio fx-9860G AUOnce the graph is drawn, use the G-Solv function to selectMIN to find the turning point.

Once again use the G-Solv function to select ROOT to

find the x-intercepts. To find the other zero, press .

The minimum is at (2.5, −2.25) and the x-intercepts are 1 and 4.

Texas Instruments TI-84Once the graph is drawn, press to enter the

CALC (calculate) menu. Select 3: minimum.

The calculator will ask for a left bound, right bound and guess.

Since the curve is simple, press for each question. The calculator will display the coordinates of the minimum.

To find a zero, select 2: zero from the CALC menu.

When the calculator asks for the left bound, use to

place the cursor before the zero and press .

For the right bound, use to place the cursor after the zero

and press . When it asks for guess, just press

or place the cursor close to the zero first. The calculator will then display the coordinates of the zero. Repeat to find the other zero.

The minimum is at (2.5, −2.25) and the x-intercepts are 1 and 4.


WINDOW

2nd TRACE

ENTER

ENTER

ENTER ENTER

Example 17




Exercise 6.3 Sketching quadratic functions1 Find the position and type of turning point for each of the quadratic functions graphed below.

a b

c d

2 Use the turning point and y-intercept to determine the equation of each of the functions shown in question 1.

3 Which of the following graphs best represents the quadratic function y = −x2 − 2x − 3?


1 2 3 4 5 6 7−1 x

2

4

−6

−4

−8

−10

−12

y

12

14

6

8

4

2

−2

y

−3 −2 −1−4 x

10

3

3.5

1.5

2

1

0.5

−0.5

y

−3 −2 −1 x

−1

2.5

10

12

4

6

2

−2

−4

y

1 2 3−1 x

8

−1−2 1 3 x−3 2

y

2

−4

−6

−1−2 1 3 x−3 2

y

2

−4

−6

A B

−1−2 1 3 x−3 2

y

2

−4

−6

−1−2 1 3 x−3 2

y

2

−4

−6

C D


Parabolic functions

240 New QMaths 11B9780170136570

4 Match each of the following quadratic functions with the correct sketch.a f (x) = x2 + 5 b f (x) = −x2 + 8x c f (x) = −3x2 + 4x + 4d f (x) = 2x2 − x − 3 e f (x) = 2x2 + 9x + 5 f f (x) = −6x2 + 7x + 3

5 State the position and type of turning point for each of the following quadratic functions.a y = (x + 1)2 − 2 b y = 3(x − 2)2 + 3 c y = 9 + (x − 1)2

d y = 1 − 2(x + 3)2 e y = 5[12 − 4(x + 2)2] f y = −2[1 + 2(x − 4)2]

6 Find the position and type of turning point for each of the following quadratic functions.a y = x2 − 3x + 1 b y = −x2 + 7x + 2 c y = x2 − 5x + 2d y = x2 − 3x − 18 e y = x2 − 7x + 5 f y = 3x + 1 − x2

7 Find the position and type of turning point for each of the following quadratic functions.a f (x) = x2 + 6x + 10 b f (x) = 2x2 + 16x + 28 c f (x) = 10x − x2 − 25d f (x) = 16x2 + 628 − 208x e f (x) = 4x2 − 20x + 41 f f (x) = −3x2 − 24x − 57g f (x) = x2 − 4x − 12 h f (x) = −x2 − 9x − 14

8 For the quadratic function y = x2 − 4x − 12 find the:a y-intercept b x-intercepts (if any)c nature and position of the turning point d axis of symmetry.Sketch the graph and state the domain and range.

9 For the quadratic function y = −2x2 − 2x + 24 find the:a y-intercept b x-intercepts (if any)c nature and position of the turning point d axis of symmetry.Sketch the graph and state the domain and range.

10 Sketch the following quadratic functions, showing all key points, and state the ranges.a y = (x + 3)2 − 4 b y = −x2 − 3x, for −4 � x � 2c y = x2 − 2x − 2 d y = x2 − 9, for −4 � x � 4e y = 9 − x2 f y = 9 − 6x − 3x2, for −2 � x � 4g y = −3x(x − 4) h y = 4x2 − 4x − 3i y = 8 − 10x − 3x2, for −5 � x � 2 j y = −3x2 + 12xk y = 16x − 4x2 l y = 3x2 − 12x − 15

A Bf(x)6

2

−1 2 x

−4

1 3

f(x)

4

2

−1 2 x

−4

1 3

f(x)64

−2 2 x

−4

−4

−6

24

C

D E f(x)

4

2

2 x1

6

−2

2 x−2

10

f(x)

2

4

6

8

0

20f(x)

−4

161284

2 4 6 8 x−2

−4

F

−1

31−1−3



11 Use your graphics calculator to find the key points of:a y = x2 + 6x + 2 b f (x) = 9 − 2x2 − 5xc y = 23x − 4x2 − 1 d f (x) = 7 − 5x + 2x2

12 Find the equations of the following parabolas in the form f (x) = ax2 + bx + c.a y-intercept of −6 and turning point at (2, −2)b y-intercept of −3 and x-intercepts of −1 and 2c x-intercepts of −2 and 3 and a maximum of 5

Modelling and problem solving13 Show that the turning point of y = ax2 + bx + c occurs at by completing

the square.

14 Find the distance between the turning points of y = 10x − x2 and 2y = 75 + 10x − x2.

15 The water solubility of a particular compound changes with temperature as follows.Maximum amount dissolved (g/L) = 25 − T + 0.0125T 2 where T is measured in °C.

Sketch a graph of this relationship, and hence find the temperature with the minimum solubility.

16 The diagram on the right shows a square plot of ground with a side length of 40 m. Two areas of the plot, �PAQ and �CDQ, are found to be unsuitable for planting crops. The plot owner wants to used the remaining (shaded) area for planting vegetables. Show that the area, A, available for planting vegetables is given by

A = 800 + 20x − x2

and find the value of x for which the area is a maximum.

17 Find the distance from the turning point of 20y = x(50 − x) to the line y = 40.

18 The height of a ball thrown upwards at 25 m/s is approximately given by the equation

h(t) = 2 − 5t2 + 25tFind the turning point and intercepts, and hence the time taken to reach the maximum height and the time taken to reach the ground.

6.4 Finding simultaneous solutionsYou have already seen that simultaneous linear equations can be solved graphically by finding the intersection of the lines. A similar method can be applied to simultaneous equations where one is linear and the other quadratic.

b–2a------ 4ac b2–

4a---------------------,⎝ ⎠

⎛ ⎞

A

B C

D

P

Q

x

x

12---


Parabolic functions

242 New QMaths 11B9780170136570

Use graphs to find simultaneous solutions to y = 2x2 − x − 10 and y = 4 − 4x.

SolutionReasonably accurate graphs are needed for the solution, so are best drawn on graph paper.Graph 1

It is useful to add some extra points to help sketch the graphs.

Graph 2

Plot both graphs on the one set of axes.

From the graph, it appears that the intersections are at about (−3.5, 18) and (2, −4).

Write the equation. y = 2x2 − x − 10Identify the features of the graph. It is a parabola with y-intercept −10.Let y = 0 to find zeros. 0 = 2x2 − x − 10Factorise. 0 = (2x − 5)(x + 2)Solve. 0 = (2x − 5) or 0 = (x + 2)

x = 2 or x = −2

State the result. Zeros exist at −2 and 2 .

Write the equation. y = 2x2 − x − 10

Express so x2 term has coefficient of 1. y = 2 − 10

Complete the square. y = 2 − 10

Separate out the perfect square. y = 2 + 2 × − − 10

Factorise the perfect square and simplify. y = 2(x − )2 − 10

State the turning point. Minimum is at ( , −10 ).

Write the equation. y = 4 − 4xLet y = 0. x-intercept is 1.Let x = 0. y-intercept is 4.

The exactness of these solutions can be checked by substitution into the equations.

2(−3.5)2 − (−3.5) − 10 = 18 OK4 − 4(2) = −4 OK

Write the answer. The simultaneous solutions arex = −3.5, y = 18 and x = 2, y = −4.

12---

12---

x2 x2---–⎝ ⎠

⎛ ⎞

x2 x2---– 1

16------ 1

16------–+⎝ ⎠

⎛ ⎞

x2 x2---– 1

16------+⎝ ⎠

⎛ ⎞ 116------

14--- 1

8---

14--- 1

8---

x −4 −1 4

y 26 −7 18

−5

−10

−2−3

10

5

15

20

25y

x1 32−1

y = 2x2 − x − 10y = 4 − 4x

Example 18



Three possible cases exist for the solution of simultaneous equations with a quadratic and a linear equation:■ two solutions (as in Example 18)■ one solution, where the line is a tangent■ no solutions, where the lines do not intersect.

A graphics calculator is particularly useful whensolving this type of problem.

Onesolution

Linearfunctions

Quadraticfunction

Two

No solutions

y

x

solutions

Use a graphics calculator to solve y = x2 + 2x − 3 and x − 2y − 4 = 0.

SolutionFor all calculatorsBoth functions must be in the form y = f(x), so we change the second function to y = − 2.Enter the functions into Y1 and Y2 respectively.Set the (or V-Window) to a domain of [−4, 2] and a range of [−5, 1] and draw the graphs.

Casio fx-9860G AUOnce the graphs are drawn, use the G-Solv function to select ISCT to find the points of intersection of the graphs.

The first point of intersection will then be displayed.

To find the other point of intersection, press . The second point of intersection will then be displayed.

The intersections are at (−2, −3) and (0.5, −1.75), so the simultaneous solutions are x = −2, y = −3 and x = 0.5, y = −1.75.

Texas Instruments TI-84Once the graphs are drawn, press to enterthe CALC (calculate) menu. Select 5: intersect.The calculator will ask for the first curve. The pointer willalready be positioned on the first curve, so just press .

Press again when prompted for the second curve.

x2---

WINDOW

2nd TRACE

ENTER

ENTER

Example 19


Parabolic functions

244 New QMaths 11B9780170136570

There are times when an exact solution of simultaneous quadratic and linear equations is required. In these cases, the equations can be solved algebraically by substitution. The linear equation is changed to a formula for y and substituted into the quadratic. If necessary, the quadratic formula may be used to find the values of x. These are then substituted back into the linear equation to find the corresponding y values.

The calculator will display Guess? Move the cursor near the first intersection and press . The intersection will then be shown at the bottom of the screen.

Repeat to find the second intersection.

The intersections are at (−2, −3) and (0.5, −1.75), so the simultaneous solutions are x = −2, y = −3 and x = 0.5, y = −1.75.


ENTER


Solve y = x2 − 3x + 2 and 2x + y − 5 = 0 algebraically.

SolutionWrite down the linear equation. 2x + y − 5 = 0Rearrange to make a formula. y = 5 − 2xWrite down the quadratic equation. y = x2 − 3x + 2Substitute the formula for y into the quadratic. 5 − 2x = x2 − 3x + 2Simplify and regroup terms on the LHS. x2 − x − 3 = 0The resulting quadratic will not factorise, so write down the quadratic formula.

x =

Substitute a = 1, b = −1 and c = −3. =

Simplify and find approximate solutionsfor x.

= ≈ 2.30 or −1.30

Write down the linear formula. y = 5 − 2xSubstitute the exact values of xinto the linear formula.

= 5 − 2

Expand each solution separately. = 5 − (1 + ) or 5 − (1 − )

Simplify and find approximatesolutions for y.

= 4 − or 4 + ≈ 0.39 or 7.61

State the solution. The solutions are approximately(2.30, 0.39) and (−1.30, 7.61).

b– b2 4ac–±2a

----------------------------------------

1 12 4 1 3–××––±2 1×

-------------------------------------------------------

1 13±2

--------------------

1 13±2

--------------------⎝ ⎠⎛ ⎞

13 13

13 13

Example 20



Exercise 6.4 Finding simultaneous solutions1 Use the graph to estimate the simultaneous solutions for each situation shown below correct

to 1 decimal place.

2 Solve the following simultaneous equations graphically.a y = 2x2 + 4x − 13 and y = 3x + 2 b y = x2 + 5x − 9 and 4x − y − 3 = 0c y = 4x2 − 10x + 4 and y = 2x + 5 d y = x2 − 4x − 5 and x − y + 2 = 0

3 Use a graphics calculator to solve the following.a y = 2x2 + 4x + 2 and x + y − 14 = 0 b y = 9x2 − 20x + 15 and x − 0.1y − 1 = 0c y = x2 + 4x + 4 and y = x − 4 d y = x2 + 2x − 4 and x + y = 1

4 Solve the following simultaneous equations algebraically, expressing answers in exact form and correct to 2 decimal places where appropriate.a y = 4x2 + 10x + 30 and 10x + y = 5 b y = 2x2 + 4x + 5 and y = x + 1c y = 2x2 − 4x − 10 and y = x + 2 d y = x2 − 4x + 6 and x − y + 2 = 0e y = x2 − x − 8 and y = 2x − 3 f y = 2x2 − 10x + 20 and x − 0.2y + 1 = 0g y = 3x2 + 3 and x − y + 1 = 0 h y = x2 + 8x + 6 and y = 2x + 1

5 The linear function y = 2x + 7 and quadratic function y = 2x2 − 5x − 8 intersect at P and Q as shown in the graph on the right. Find the simultaneous solutions of the functions, and hence calculate the length of PQ.

Modelling and problem solving6 Find the simultaneous solutions of the functions

y = 2x + 4 − x2 and y = 2x2 − x − 2.


a b16

−12−8

1284

y

16

246

141210

y

4−6 2−4 x−2 −4

−3 1−2 x−1

18

8

−4

−2 6 x2 4−2

8

−8−6−4

642

y10

2−2 x31−1−2

−14−12−10

8

−8−6−4

642

y10

−12−10

c d

−16

−38

20

6−4

−15−10−5

15105

42−2

y

x

y = 2x2 − 5x − 8

y = 2x + 7 Q

P


Chapter summary

246 New QMaths 11B9780170136570

■ A polynomial function in x is defined as:f(x) = anxn + an − 1xn − 1 + an − 2xn − 2 + … + a1x1 + a0

where n is known as the degree of a polynomial.

■ A quadratic function (or quadratic) is a polynomial of degree 2. The standard form of a quadratic function is f(x) = ax2 + bx + c (a ≠ 0). The standard form of the corresponding quadratic equation is ax2 + bx + c = 0.

■ Quadratic equations have 0, 1 or 2 solutions. The solutions of a quadratic equation are roots or zeros of the corresponding quadratic function.

■ Quadratic equations may be solved by numerical methods such as systematic trial and error. Solutions may also be found by graphing.

■ Algebraic methods for the solution of quadratic equations use the null factor law:If ab = 0, then either a = 0, b = 0 or a = b = 0.

■ Algebraic methods include factorisation, completion of the square and the quadratic formula.

■ The quadratic formula for the equation ax2 + bx + c = 0 is x = .

■ The graph of a quadratic function has a distinctive parabolic shape. Hence, quadratics are also known as parabolic functions.

■ Like all parabolas, the graph of a quadratic function is symmetrical about an axis drawn through the turning point—the point at which the graph changes slope from positive to negative or vice versa.

■ The turning point is a minimum if the quadratic has a positive coefficient of the x2 term, or a maximum if the coefficient is negative.

■ For the graph of a quadratic, the domain is the set of real numbers. The range must be greater than or equal to the minimum value, or less than or equal to the maximum value.

■ The important features of the graph of a quadratic function are:– the turning point– the zeros– the y-intercept.

■ The turning point of a quadratic function can be found by completing the square. The process changes y = ax2 + bx + c to y = a(x − p)2 + q to give the turning point (p, q). Alternatively, the turning point of the quadratic functionf(x) = ax2 + bx + c is

■ Simultaneous equations that involve a quadratic may have no, one or two solutions. The solutions can be found graphically by finding points of intersection or algebraically by substitution.

b– b2 4ac–±2a

-------------------------------------

ZeroMinimum Maximum

Axi

s

Axi

s

Zero x

y

−b2a------ , f −b

2a------⎝ ⎠

⎛ ⎞⎝ ⎠⎛ ⎞

Two

One

No solutions

y

x

Quadratic

Linear

function

functions

solutions

solution


Chapter review


Knowledge and procedures1 Estimate the roots of the following quadratic functions correct to 1 decimal place.

2 A solution to 4x2 + x − 5 = 0 is:A −5 B −1.25 C 0.986 D 1.25 E −0.986

3 An approximate solution of x2 − 5x − 7 = 0 is:A 12.28 B 2.28 C −6.14 D 6.14 E −2.28

4 Use systematic trial and error to find solutions to:a 4x2 + x = 3 b 2x = 3x2 − 6

5 Find solutions to 2x2 = 5x + 12 by systematic trial and error. You may use a spreadsheet.

6 An approximate solution of 2x2 + 5x = 8 is:A 3.61 B 1.11 C 0.89 D −0.89 E 2.54

7 Plot a graph to find solutions to 3x2 + 4x − 15 = 0.

8 Factorise each of the following quadratics.a 2x2 − x − 6 b 3x2 + 2x − 8 c 10x2 − 19x − 15

9 The equation 3x2 − x − 4 = 0 factorises to:A (3x + 1)(x − 4) = 0 B (3x − 1)(x + 4) = 0 C (3x + 4)(x − 1) = 0D (3x − 4)(x + 1) = 0 E (3x + 2)(3x − 2) = 0

10 By completion of the square, the equation 2x2 + 12x − 3 = 0 is changed to:

A (x − 6)2 = 1 B (x + 3)2 = 12 C (x + 3)2 = 10

D (x − 3)2 = −12 E (x + 6)2 = 37

11 Solve by completion of the square: x2 + 2 = 8x.

12 Completion of the square changes 2x2 + 4x − 5 = 0 to:A (x + 2)2 = 9 B (x + 1)2 = 6 C (x + 1)2 = 3.5D (x + 2)2 = 6.5 E (x + 1)2 = −3.5

13 Solve by factorisation: 6x2 − 17x = 14.

14 Solve the following using the quadratic formula, giving answers both in exact form and correct to 2 decimal places.a 2x2 + 7x + 1 = 0 b x2 + x = 3

Ex 6.1

8

123

765

y

1 42 x3

9

4

−1

a b

−2

−14−12

−4−6−8

y2

−10

−2 1 42 x3−1

Ex 6.1

Ex 6.1

Ex 6.1

Ex 6.1

Ex 6.1

Ex 6.1

Ex 6.2

Ex 6.2

Ex 6.2

12--- 1

2---

12---

Ex 6.2

Ex 6.2

Ex 6.2

Ex 6.2


Chapter review

248 New QMaths 11B9780170136570

15 A toy bought for $x is sold at a loss of x% for $16. Write an equation and find the cost of the toy.

16 The solutions of 3x2 + 4x − 5 = 0, correct to 2 decimal places, are:A 1.77 and −0.44 B −1.77 and 0.44 C 2.12 and −0.79D −2.12 and 0.79 E −6.36 and 2.36

17 Use the quadratic formula to solve, correct to 2 decimal places:a x2 + 3x − 1 = 0 b 3x2 + 9x = 11c 5x + 10 = 2x2 d x2 = x + 8

18 Find the position (correct to 1 decimal place) and type of turning point for each of these quadratic functions.

19 The turning point of f(x) = 6x − x2 + 9 is:A a minimum at (3, 9) B a maximum at (3, 18) C a minimum at (3, −18)D a maximum at (−3, 18) E a minimum at (−3, 18).

20 The function y = 10x − x2 − 4 has:A a minimum at (5, 21) B a maximum at (5, 21) C a minimum at (10, −4)D a maximum at (10, −4) E a minimum at (20, −204).

21 Complete the square to find the turning point of:a y = x2 − 6x + 4 b f (x) = x2 + 4x + 8c f (x) = 8x − x2 − 19 d y = x2 + 2x − 1

22 The x-intercepts of f(x) = 3x2 − 15x − 18 occur at:A 6 and −1 B −6 and 1 C 6 and −3D −6 and 3 E −6 and −1.

23 For y = 5x − 2x2 + 10, find the position and type of the turning point. Sketch the graph.

24 Use a graphics calculator to draw the graph of P(x) = x2 + 5x − 6. Sketch what you see and show the connection between the zeros of P(x) = x2 + 5x − 6 and the factors of the expression x2 + 5x − 6.

25 Sketch each of the following quadratic functions.a f (x) = x2 − 10x + 21 b y = 15 − 2x − x2

c y = x2 − 3x − 4 d f (x) = 6x − x2 − 4

Ex 6.2

Ex 6.2

Ex 6.2

Ex 6.3

8

123

765

−2 −1 x1

9

4

−3

a b

−2

−14−12

−4−6−8

y

2

−10

1 42 x3−1

4y

−1

Ex 6.3

Ex 6.3

Ex 6.3

Ex 6.3

Ex 6.3

Ex 6.3

Ex 6.3


Chapter review


26 State the range of each of the following functions over the given domain.a y = x2 + 2x − 5 for −2 � x � 3 b y = 4x − x2 for −1 � x � 5c f (x) = −x2 − 14 − 6x for −5 � x � 2 d y = x2 − 6x + 11 for 1 � x � 7

27 Use the graph to estimate the simultaneous solutions for each situation shown below correct to 1 decimal place.

28 The intersections of y = x2 − 2x − 8 and y = x + 2 occur at:A (4, 6) and (2, 4) B (4, 6) and (−2, 0) C (−4, −2) and (2, 4)D (5, 7) and (−2, 0) E (−5, −3) and (2, 4)

29 Solve graphically: y = x2 + x − 1 and y = 3x + 7.

30 Solve algebraically: y = 2x2 − x − 3 = 0 and x − 0.5y + 1 = 0.

31 Find the solutions to the following simultaneous equations by graphing.a y = x2 − 6x + 10 and y = 2x − 5 b y = 8 − 2x2 and y = x − 2c 3x − y − 2 = 0 and y = x + 6 − 3x2 d y = 6x2 + 9x − 3 and 2x + y = 7

32 Find the solutions to the following simultaneous equations algebraically.a y = 3x2 + 9x − 6 and x + y − 2 = 0 b y = 6 − x2 − 5x and y = 1 − xc y = 3x2 − 3x − 3 and y − 4 − 2x = 0 d y = 1 + 2x − 3x2 and y + 4 = 3x

Modelling and problem solving33 A golden rectangle has sides in the golden ratio x : 1 such that, if a square is cut off, the ratio

of the sides of the remaining rectangle is also x : 1.

a Find two ways to write the length of y in terms of x.b Hence write an equation involving only x.c Rearrange the equation to standard form.d Solve the equation to find the golden ratio x.e If one side of a golden rectangle is to be of length 5 m, what are the possible lengths of

the remaining side?

Ex 6.3

Ex 6.4

a by

1 42 x3

1

4

−8−6

2

−2

y

−4

1 42 x3

86

−2

−8−7

−3−4−5−6

−1−1

−1

Ex 6.4

Ex 6.4

Ex 6.4

Ex 6.4

Ex 6.4

Ex 6.2

11

yxx:1 = 1: y


Chapter review

250 New QMaths 11B9780170136570

34 The sketch shows the path of a ball thrown from atreehouse. The ball reaches a maximum height of10 m after being thrown from a height of 5 m. Themaximum height is reached a distance of 20 m fromthe treehouse, and the ball is again level with thetreehouse a horizontal distance of 40 m from thetreehouse.

a Place the information in a table, with the distance from the treehouse as x and the height as y.

b Find a polynomial expression for the relationship between x and y.c The horizontal distance may be written as x = kt where t is the time taken. If the time

taken to reach the greatest height is 0.6 s, find the value of k.d Write the vertical distance as a function of t.e How far from the treehouse does the ball strike the ground?

35 An open box is 6 cm high, and its length is to be 4 cm more than its width.a Write the volume V in terms of the width w.b Write the surface area S in terms of the width w.c Draw graphs of V and S on the same set of axes for values of w from 1 to 6 cm.d Find the value of w for which V and S have the same numerical value.e What happens to the ratio of V to S as w increases?

10 m

20 m 20 m

5 m

Ex 6.3

Ex 6.4


00 NQM11B SB TXT.prelim Q Maths 11B/ch.06.text.pdf · Solving quadratic equations 6.2 Algebraic solution of quadratics 6.3 Sketching quadratic functions 6.4 Finding simultaneous solutions

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