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Quadratic Function

Nov 20, 2014

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Complete lesson on quadratic functions was originally a notebook file but compressed to pdf
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Page 1: Quadratic Function

1

Chapter 10

Page 2: Quadratic Function

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What I already know about a quadratic equation..

Page 3: Quadratic Function

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Everything you always wanted to know about the parabola

Drag the correct label

Axis of Symmetry

Root

y intercept

Vertex

Page 4: Quadratic Function

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How do we find the X Intercept?

from the graph

from the equation y = 4x2-3x - 1

LOOK

The roots of an equation are the points where the graph cuts the x axis.They are also the x intercepts.

put y = 0, factorise and solve

Page 5: Quadratic Function

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How do we find the X Intercept?

from the graph

from the equation

LOOK

put y = 0, factorise and solve

y = -3x2-5x + 2

The roots of an equation are the points where the graph cuts the x axis.They are also the x intercepts.

Page 6: Quadratic Function

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How do we find the Y Intercept?

from the graph

from the equation

LOOK

put x = 0 and find y

y = -3x2-5x + 15

The y intercept is the point where the graph cuts the y axis

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The general equation for any parabola is y = ax2 +bx + c

a = coefficient of x2b = coefficient of xc = constant

coefficient just means number in front of

What happens to the shape / position of the parabola when:a is changed?

c is changed?

b is changed?

Website link: http://tinyurl.com/34tkg7z

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Axis of symmetry is the average of the x intercepts

How do we find the axis of symmetry?

from the graph

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If a quadratic cannot be factorised we can solve it by using the quadratic formula

y = ax2 +bx + c

-b + !b2 - 4ac2a

x =

-b + !b2 - 4ac2a

-b - !b2 - 4ac2a

Axis of symmetry is the average of the x intercepts

How do we find the axis of symmetry?

from the equation

x = x =

-b 2a

x =

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example: find the axis of symmetry of y = 4x2-3x - 1

The axis of symmetry is a vertical line (parallel to the y axis) and is given by the equation x = -b

2a

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from the graph LOOKMAX

MIN

How do we find the vertex?

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eg. find the vertex of y = 4x2-3x + 1

How do we find the vertex?

Find the axis of symmetry x = - b 2aand substitute this value of x into the equation to find the y coordinate.

Where is the vertex in relation to the axis of symmetry?

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Match the picture with its description:

y = 2x2 + 8x - 3

axis of symmetryroots

parabola

quadratic equation

vertex

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Questions:

1. Sketch the graph of the parabolas which have roots 3 and -1

2. What is the equation of each graph?

3. Is the axis of symmetry the same for each graph?

4. What is it?

5. Find the maximum and minimum for each graph.

Page 15: Quadratic Function

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Questions:

1. Sketch the graph of the parabola y= x2 - 3x

2. What is the axis of symmetry?

3. Does the parabola have a maximum or minimum value? Why?

4. What is this value?

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Solving quadratic inequations

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-1 0-2-3-4-5 1 2 3 4 5

-1 0-2-3-4-5 1 2 3 4 5

-1 0-2-3-4-5 1 2 3 4 5

-1 0-2-3-4-5 1 2 3 4 5

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Solving Quadratic Inequations

y = ( x - 5 ) ( x + 2 )

( x - 5 ) ( x + 2 ) > 0

( x - 5 ) ( x + 2 ) < 0

Page 19: Quadratic Function

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Solving Quadratic Inequations

y = ( x + 5 ) ( 2 - x )

( x + 5 ) ( 2 - x ) " 0

( x + 5 ) ( 2 - x ) # 0

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Positive and Negative Definitesor quadratic equations with no solutions

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How do we find the axis of symmetry?

What are the solutions (roots)?

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Solving Quadratic Inequations

y = x2- 4x + 9

x2- 4x + 9 > 0

x2- 4x + 9 < 0

Page 23: Quadratic Function

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Solving Quadratic Inequations

y = - x2+ 4x - 5

-x2+ 4x - 5 > 0

-x2+ 4x - 5 < 0

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Positive Definite Quadratic Function Negative Definite Quadratic Function

function is always > 0 (above x axis)

function is always < 0 (below x axis)

These quadratics have no solutions ( roots) because they do not touch the x axis.

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Indefinite Quadratic functions

indefinite functions because they are sometimes > 0 (above x axis) and sometimes < 0 ( below the x axis)

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6 maximum point

2 minimum point

3 quadratic formula

7 indefinite

5 positive definite

1 parabola

4 negative definite

The graph of a quadratic function is always a ....

A quadratic with no solutions where a > 0

A quadratic with no solutions where a < 0

The quadratic y = ax2+bx + c

where a > 0 will have a ....

The quadratic y = ax2+bx + c

where a < 0 will have a .....

The quadratic y = ax2+bx + c

which has two roots is called ...

Can be used to find the roots of y = ax2 + bx + c

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How many solutions?

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How many roots are there for the equation: y = x2- 4x - 5.What are they?Sketch the graph.

x intercepts:y = (x - 5) (x + 1)

(x - 5) (x + 1) = 0

x = 5 and x = -1 two solutions

Axis of symmetry x = 4/2 = 2when x = 2 y = -9

coefficient of x2 is positiveconcave up in shape.

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How many roots are there for the equation: y = x2- 4x + 4.What are they?Sketch the graph.

y = (x - 2)2

(x - 2)2 = 0x = 2 only 1 solution

Axis of symmetry x = 4/2 = 2when x = 2 y = 0

coefficient of x2 is positivegraph is concave up

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How many roots are there for the equation: y = x2- 4x + 5.Sketch the graph.

y = x2- 4x + 5

axis of symmetry x = 4/2 = 2

when x = 2 y = 1

a >0 (coefficient of x2 ).The graph has a minimum point at (2,1)

hence the graph is positive definite ( sits above the x axis)

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-b + !b2 - 4ac2a

x =

y = 4x2- 3x - 5

3 + !(-3)2 - 4.4.-58

x =

3 + !9 +808

x =

3 + !898

x =

3 + !898 x = 3 - !89

8 x =

Find the roots of:

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-b + !b2 - 4ac2a

x =

y = x2 - 4x + 4

4 + !(-4)2 - 4.1.42

x =

4 + !16 - 162

x =

4 2

x =

Find the roots of:

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-b + !b2 - 4ac2a

x =

y = 2x2 - 3x + 5

3 + !(-3)2 - 4.2.54

x =

3 + !9 - 404

x =

Find the roots of:

No soln.

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Page 35: Quadratic Function

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The solutions (roots) for the general quadratic is y = ax2 +bx + c

coefficient just means number in front of

-b + !b2 - 4ac2a

x =

-b + !b2 - 4ac2a

-b - !b2 - 4ac2a

x = or 2 solutions

-b + 02a

x = 1 solution

No solution -b + !negative number2a

x =

Page 36: Quadratic Function

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Discriminant

= b2 - 4ac

The discriminant gives information about the nature of solutions for the given quadratic equation such as how many solutions (roots) and the type of solution .

< 0 = 0 > 02 real solutionsNo real solutions

(unreal / imaginary)

a < 0negative definite

a > 0positive definite

1 real (rational solution)equal roots

perfect square2 rational solutions 2 irrational solutions

not perfect square

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Examples

1. Find the discriminant of 3x2 - 5x + 5 = 0 and determine whether the roots are real / unreal (imaginary), rational / irrational, equal / unequal?

= b2 - 4ac

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Examples

2. Find k if 3x2 - 8x + k = 0 has equal roots

= b2 - 4ac

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Examples

3. Show that 4x2 + 3kx + k

2 = 0 has no real roots whatever the value of k

= b2 - 4ac

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Summary

= b2 - 4ac

= 0 equal roots

< 0 unreal (imaginary roots)

! 0 real roots

> 0 real and different

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Intersection of Curves and Lines

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To find the point of intersection, the two equations are solved simultaneously.

We only have to show how many solutions.Look at the discriminant when one pronumeral has been eliminated.The discriminant shows how many solutions -

eg. The line y = 3x - p + 1 is a tangent to the parabola y = x2. Find the value of p.

y = 3x - p + 1 y = x2 3x - p + 1 = x2 ie x2 -3x + p - 1 = 0

The discriminant of this quadratic = (-3)2 - 4.1.(p-1) = 9 - 4p + 4 = 13 - 4p

We need to show the line is a tangent which means there is only one point of intersection. This occurs when the discriminant is zero.

Solve 13 - 4p = 0 gives p = 13/4 = 3.25

> 0 then there are 2 solutions ( 2 points of intersection)= 0 then there is 1 solution (tangent) < 0 then there are no points of intersection

}

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If two quadratic expressions are equivalent to each other then the corresponding coefficients must be equal.

eg. Express x2 +2x - 2 in the form ax(x+1) + bx2 +c(x+1)

x2 +2x - 2 = ax(x+1) + bx2 +c(x+1)

in other words find the values of a, b and c so that the expressions are equal.

answer: x2 +2x - 2 = 4x(x+1) - 3x2 -2(x+1)

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If two quadratic expressions are equivalent to each other then the corresponding coefficients must be equal.

eg. Find the equation of the parabola which passes through the points (2,10) (1,1) and (0,-4)

As these points lie on the same parabola, then we need to find a, b and c so that ax2 + bx + c = y

substitute each point into the general quadratic equation and solve for a, b and c

Required equation of the parabola is 2x2+3x -4 = y

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Sum and Product of the Roots of a Quadratic Function

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x2 - 5x + 6 = 0

Tell me everything you know about:

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If ! and β are the roots of the general quadratic then

x2 - ( ! + β) x + !β = 0

ie. x2 - ( sum of the roots) x + product of the roots = 0

Example: The roots of a quadratic are 3 and -2. What is the equation of the quadratic?

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If ! and β are the roots of the ax2 + bx + c = 0

sum of the roots = ! + β = -b a

ca

product of the roots = !β=

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Examples: For the quadratic 2x2 - 5x +1 = 0 find;

a) ! + β

b) !β

c) 1 + 1! β

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Examples: For the quadratic 2x2 -5x +1 = 0 find; a = 2 , b = - 5, c = 1

d) ! 2 + β2

e) (! - 3 ) ( β- 3 )

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Example 2: Find k in x2 - (2 + k)x + 3k = 0 where the product of the roots is 4 times the sum of the roots.

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Example 3: Find the value of k if the roots of 4x2 - 20x + k = 0 differ by 2.

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Equations Reducible to Quadratics

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1. By multiplying every term by the denominator

Solve: m + 1 = 2 m

Some equations can be solved by making them into a quadratic ax2 + bx + c = 0.

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2. By making a substitution

Solve: x4 -5x2 + 4 = 0

Solve: (2x - 1 )4 - (2x - 1 )2 - 6 = 0

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2. By making a substitution

Solve: 9x - 10(3x)+ 9 = 0