GENERAL to Transformational CHANGING FORMS. To General Form – Complete the squar Remember: the general form of a quadratic function is y = ax 2 + bx.

GENERAL to Transformational

CHANGING FORMS

To General Form – Complete the squar Remember: the general form of a

quadratic function is y = ax2 + bx +c The transformational form is The standard form is y = a(x-h)2 + k To GET general form: FOIL and solve for

y.

2)()(1

hxkya

But how do we go FROM general form?

To General Form – Complete the squar

Let’s look at an example:

What is the vertex of the quadratic function y = 2x2 + 12x -4?

We know that in transformational form the coefficient of ‘x’ is 1

2)()(1

hxkya

So STEP 1 is to divide every term by ‘a’

262

1 2 xxy

STEP 2 is to move the non-x term over.

xxy 622

1 2

Completing the Squar

xxy 622

1 2

The right-hand side of the equation needs to be a perfect square: (x –h)2

The right-hand side of the equation needs to be a perfect square: (x –h)2

So far we have: x2 +6xThis looks like:So far we have: x2 +6xThis looks like:

Completing the Squar

xxy 622

1 2

To make this into a perfect square we could…To make this into a perfect square we could…

Completing the Squar

xxy 622

1 2

To make this into a perfect square we could…To make this into a perfect square we could…

Almost…we need to complete the square!

Almost…we need to complete the square!

We were missing nine 1x1 squares.We were missing nine 1x1 squares.

Completing the Squar

xxy 622

1 2 So the PERFECT SQUARE is …So the PERFECT SQUARE is …

---------- X + 3 -----------

---------- X +

3

-----------

(x+3)2

(x+3)2

xxy 622

1 2

222 36322

1 xxy

We took HALF the coefficient of x and squared it.

We took HALF the coefficient of x and squared it.

Added it to the right hand side. This means we must add it to both sides!

Added it to the right hand side. This means we must add it to both sides!

2)3(112

1 xy

STEP 3: take half the coefficient of x, square it and add it to both sides

Completing the Square

xxy 622

1 2 So the PERFECT SQUARE is …So the PERFECT SQUARE is …

---------- X + 3 -----------

---------- X +

3

-----------

(x+3)2

(x+3)2

xxy 622

1 2

222 36322

1 xxy

2)3(112

1 xy

We need the left hand side to have brackets.

2)3()22(2

1 xy

Step 4: Factor out the 1/a term on the left hand side

Step 4: Factor out the 1/a term on the left hand side

Completing the square

We started with the quadratic in general form y = 2x2 + 12x -4. The same function is

in transformational form.

2)3()22(2

1 xy

So what is the range of y = 2x2 + 12x -4 ?

Since the equivalent equation shows us a vertex of (-3,-22) with no reflection, the range is

Since the equivalent equation shows us a vertex of (-3,-22) with no reflection, the range is

),22[ y

Practice going FROM general form1. What is the vertex of the function y =

2x2 - 8x +2?

2. What is the range of the function y = -x2 -5x +1?

3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

Practice going FROM general form1. What is the vertex of the function y = 2x2 - 8x

+2?14

2

1 2 xxy

xxy 412

1 2

44412

1 2 xxy

2)2(32

1 xy

2)2()6(2

1 xy

Vertex: (2,-6)

Practice going FROM general form2. What is the range of the function y = -x2 -

5x +1?152 xxy

xxy 51 2

222 5.255.21 xxy

2)5.2(25.7 xy

2)5.2()25.7( xy

]25.7,(y

Since there’s a reflection

Y-value of the vertex

Y-value of the vertex

Practice going FROM general form3. Put the equation y = 0.5x2 – 3x - 1 into

transformational form.

262 2 xxy

xxy 622 2

96922 2 xxy

2)3(112 xy

2)3(2

112

xy

A shortcut…eventually

If only there was a way to avoid having to complete the square every time to get the vertex!! Maybe there is…

Let’s put the following equations into transformational form.

y = 3x2 + 12x - 6 y = ax2 + bx + c

STEP 1 is to divide every term by ‘a’STEP 1 is to divide every term by ‘a’

243

1 2 xxya

cx

a

bxy

a 21

STEP 2 is to move the non-x term over.STEP 2 is to move the non-x term over.

xxy 423

1 2 xa

bx

a

cy

a 21

A shortcut…eventually

xxy 423

1 2 xa

bx

a

cy

a 21

22

2

22

1

a

bx

a

bx

a

b

a

cy

a222 2422

3

1 xxy

STEP 3: take half the coefficient of x, square it and add it to both sides

2)2(63

1 xy

22

22

1

a

bx

a

b

a

cy

aStep 4: Factor out the 1/a term on the left hand side

Step 4: Factor out the 1/a term on the left hand side

2)2()18(3

1 xy

22

22

1

a

bx

a

b

a

cy

a

Here’s the shortcut

We now see that ANY general equation can be written as

22

22

1

a

bx

a

b

a

cy

a

We haven’t finished this yet but we already see that the x-value of the vertex is

This is VERY important.For example: What is the axis of symmetry of the function y = 2x2 – 16x +3?

To complete the square on this takes time. But

44

16

)2(2

)16(2

x

x

x

a

bx

Finding y

Example 2: What is the range of the graph of the function y = -x2 – 4x+ 7?

We know that the x coordinate of the vertex is

22

4

)1(2

)4(2

x

x

x

a

bx

But how do I find y?

You glade the x- value!You glade the x- value!

You You

“plug in it, plug it in”“plug in it, plug it in”

11

784

7)2(4)2( 2

y

y

y

The range is The range is ]11,(y

NOTE: The vertex of ANY quadratic occurs at x = -b/2a and the max or min value is: f(-b/2a)

NOTE: The vertex of ANY quadratic occurs at x = -b/2a and the max or min value is: f(-b/2a)

Changing forms (quickly)

Ex. Graph the parabola defined by the function: f(x) = 3x2 + 12x - 9

We know that transformational form is best for graphing. Is it still necessary?

If f(x) = 3x2 + 12x – 9, then the vertex is

26

12

)3(2

122

x

x

x

a

bx

f(-2) = 3(-2)2 +12(-2)-9

f(-2) = 3*4-24-9f(-2) = -21

Vertex (-2,-21)

Changing forms (quickly)

So f(x) = 3x2 + 12x -9 can now be written: 2)2()21(1

xya

If only we knew the ‘a’ value…

2)2()21(3

1 xy

Since the VS = 3 we can graph from the vertex:Over 1 up 3Over 2 up 12Over 3 up 27

Changing Forms (quickly)

2)2()21(3

1 xy f(x) = 3x2 + 12x – 9 What we

know:

Axis of symmetry

Range (and Domain)

y-intercepty-intercept

Max/min valueMax/min value

But we don’t know the x-intercepts!But we don’t know the x-intercepts!

Finding our Roots

When dealing with quadratics, we will be asked to solve quadratic equations, find the roots or find the zeros of quadratic equations or find the x-intercepts of the parabola. These all mean: solve for x.

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15Since there are 2 x terms (an x2 and an x) we cannot undo what’s being done. So, in this form (for now) we’re stuck!

Since there are 2 x terms (an x2 and an x) we cannot undo what’s being done. So, in this form (for now) we’re stuck!

But we can change this into transformational form.

f(x) = x2 - 2x -15y = x2 - 2x -15y+15 = x2 - 2x y +15 +1 = x2 - 2x +1y +16 = (x-1)2

f(x) = x2 - 2x -15y = x2 - 2x -15y+15 = x2 - 2x y +15 +1 = x2 - 2x +1y +16 = (x-1)2

Finding our Roots

We know that for any x-intercepts: f(x) =0

y +16 = (x-1)2

f(x) = x2 - 2x -15

y +16 = (x-1)2

0+16 = (x-1)2

16 = (x-1)2 To undo what’s being done to ‘x’ we need to take the square root of both sides.

To undo what’s being done to ‘x’ we need to take the square root of both sides.

2116 x

14 x Remember: (-4)(-4) =16

Now, this is actually 2 equations in one

14 x 14 x

5

14

x

x

3

14

x

x

So the 2 x-intercepts are (5,0) and (-3,0)

So the 2 x-intercepts are (5,0) and (-3,0)

Finding our Roots

We’ve found the roots of a quadratic function but it involved us going back to transformational form. Is there a shortcut?

When we completed the square on the general form we ended up with this

22

22

1

a

bx

a

b

a

cy

a

If we set y= 0 we can come up with a formula to find the roots of a quadratic function. We’ll call it the “Quadratic Root Formula”

Finding our Roots

22

22

1

a

bx

a

b

a

cy

a

22

22)0(

1

a

bx

a

b

a

c

a

22

22

a

bx

a

b

a

c

Now, on the left-hand side we need to add fractions, so we need a common denominator.

2

2

2

24

a

bx

a

b

a

c

2

2

2

244

4

a

bx

a

b

a

c

a

a

2

2

2

24

4

a

bx

a

bac

2

2

2

24

4

a

bx

a

acb

2

2

2

24

4

a

bx

a

acb

a

bx

a

acb

24

42

2

xa

acb

a

b

2

4

2

2

Finding our Roots

xa

acb

a

b

2

4

2

2

a

acbbx

2

42

So if we go back to our original example:

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15

First set one side equal to 0 (since y = 0 for x-intercepts)

0 = x2 - 2x -150 = x2 - 2x -15

Now plug in a, b and c into the quadratic root formula

)1(2

)15)(1(4)2()2( 2 x

2

6042 x 2

642x

2

82x

52

82

x

x

32

82

x

x

So the 2 x-intercepts are (5,0) and (-3,0)

So the 2 x-intercepts are (5,0) and (-3,0)

Yet another method of finding roots

We have solved quadratic equations by putting it in transformational form and undoing what’s being done Making one side 0 and using the quadratic root formula

Sometimes we can also factor. This uses the zero property.

If (r)(s) = 0 then EITHER ‘r’ must equal 0 or ‘s’ must equal 0. This only works when the product is 0.

So if we can get the quadratic function in the form (x-r)(x-s)=0 then we know that either x-r = 0 or x –s has to be zero.

So if we can get the quadratic function in the form (x-r)(x-s)=0 then we know that either x-r = 0 or x –s has to be zero.

This is called FACTORING

Factoring to find the roots

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15

Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15

Again, f(x) = 00 = x2 -2x -15

We need two numbers whose product is -15 and whose sum is -2

Answer when the numbers are multiplied

Answer when the numbers are added

___ x ___ = -15___ + ___ = -2

So x2 -2x-15=0(x-5)(x+3)=0

So x2 -2x-15=0(x-5)(x+3)=0 Either (x-5) = 0 Or (x+3) = 0

x = 5 x = -3

So the 2 x-intercepts are (5,0) and (-3,0)

So the 2 x-intercepts are (5,0) and (-3,0)

Factoring

When factoring a quadratic equation where ‘a’ = 1, find two numbers that multiply to give ‘c’ and add to give ‘b’.When factoring a quadratic equation where ‘a’ = 1, find two numbers that multiply to give ‘c’ and add to give ‘b’.

Ex. Solve for x by factoring

24100 2 xx ___ x ___ = -24

___ + ____ = 10)2)(12(0 xx

)12(0 x or )2(0 x

12x 2x

By the way:24 = x2 +10x24 +25 = x2 +10x +2549 = (x+5)2

+/- 7 = x + 5x = -5 +7 or x = -5 -7

x = 2 or x = -12

2

14102

19610

)1(2

)24)(1(41010 2

x

x

x

x =(-10 +14)/2 =2

x = (-10-14)/2 = -12

Practice

1. Solve the following equations by factoring.

a) 0= x2 + 2x +1

b) 0= x2 + 5x +4

c) 0= x2 + 2x -24

d) 0= x2 -25

2. Find the x- and y- intercepts of the following quadratics.

a) f(x) = 2x2 +3x + 1

b) f(x) = 3x2 -7x + 2

c) f(x) = x2 -5x -14

d) y= 2(x-4)2 - 32

Answers 1a. x = -1 b. x = -4 or -1 c. x = -6 or 4 d. x = -5 or +5

2a. (-1,0) (-0.5, 0) (0,1) b. (1/3,0) (2,0) (0, 2) c. (-2,0) (7,0) (0, -14) d. (0,0) (8,0)