the powers of general form. probe below are 5 different ways of representing a quadratic...

The powers of General Form

ProbeBelow are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out.b) Name the 5 forms

x y

-1 0

0 5

1 8

2 9

3 8

4 5

5 0

2)2()9( xy

542 xxy

)9,2(),( yxyx

ProbeBelow are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out.b) Name the 5 forms

graphical (parabola)

equation in general form

equation in transformational form

table of values

mapping rule

x y

-1 0

0 5

1 8

2 9

3 8

4 5

5 0

)9,2(),( yxyx

2)2()9( xy

542 xxy

)9,2(),( yxyx

General Form toTransformational Form

Let’s look at an example.We will take the quadratic function given in general form:

y = 2x2 + 12x – 4and turn it into transformational form:

2)HT()VT(VS

1 xy

STEP 1:Divide every term by a.

262

1

4122

2

2

xxy

xxy

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

1

Let’s look at an example.We will take the quadratic function given in general form:


STEP 2:Move the non-x term over

xxy

xxy

xxy

622

1

262

1

4122

2

2

2


2)HT()VT(VS

1 xy

Completing the Squarexxy 62

2

1 2

x2 x x x x x x

We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2

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

Completing the Square

x2 x x xx xx Oops…

The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape...

x2 x x x x x xxx 62


Try again…

x2 x x x x x x

x2 x x x

x

xx

Closer….We just need tocomplete this square…

We were missingnine 1×1 squares.

xx 62


This looks promising:We have a square with length and width dimensions of (x + 3).We just had to add 9, or 32.

x2 x x x

x

xx

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

---------- x +

3

-----------222 )3(36 xxx But how can we justify

this “just add 9”?Add it to both sides of the equation!

But why 32?It’s half of the coefficient of x, squared.

962 xx

Back to our example.We will take the quadratic function given in general form:


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

2

222

2

2

2

)3(112

1

36322

1

622

1

262

1

4122

xy

xxy

xxy

xxy

xxy


2)HT()VT(VS

1 xy



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

2

2

2

2

2

2

)3(222

1

)3(112

1

96922

1

622

1

262

1

4122

xy

xy

xxy

xxy

xxy

xxy


2)HT()VT(VS

1 xy




Now we know that its VS = 2, HT = –3, and VT = –22 From this we know its vertex, range, axis of symmetry, etc

2)3()22(2

1 xy

Therefore, the transformational form of the quadratic functiony = 2x2 + 12x – 4 is:

vertex is (–3, –22)

range is{y≥ –22, y R}

axis of symmetry isx = –3

2)HT()VT(VS

1 xy

Practice

1. Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.

2. Complete the square to find the range of the functiony = –x2 – 5x +1?

3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.

142

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: Solutions

2. Complete the square to find the range of the functiony = –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

Practice: Solutions

range is{y ≤ 7.25, y R}

3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

262 2 xxy

xxy 622 2 96922 2 xxy

2)3(112 xy

2)3(2

112

xy

Practice: Solutions

A shortcut…eventuallyBut completing the square is a lot of work….

Let’s find a shortcut to finding vertex ect. by completing the square just ONE more time, but with the general equation

cbxaxy 2642 2 xxy

STEP 1:Divide every term by a.




642 2 xxy

322

1 2 xxy

A shortcut…eventuallySTEP 1:Divide every term by a.




xxy 232

1 2

222 12132

1

xxy

21132

1 xy

2142

1 xy

2182

1 xy

cbxaxy 2

a

cx

a

bxy

a 21

xa

bx

a

cy

a 21

22

2

22

1

a

bx

a

bx

a

b

a

cy

a

22

22

1

a

bx

a

b

a

cy

a22

24

1

a

bx

a

bcy

a

Here’s the shortcut

This may look complicated, but is VERY helpful…from this we can see that the HT for any general quadratic is

1

)2(2

)4(2

x

x

a

bx

For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6?To complete the square on this takes time. But…

a

b

2

We now see that ANY general quadratic equationy = ax2 + bx + c can be written as 22

24

1

a

bx

a

bcy

a


For example: What is the vertex of the function y = –2x2 + 4x + 6?To complete the square on this takes time. But…

a

bc

4

2


24

1

a

bx

a

bcy

aThis may look complicated, but is VERY helpful…from this we can see that the VT for any general quadratic is

8

)2(6

)2(4

)4(6

42

2

y

y

y

a

bcy

vertex (1, 8)

1

)2(2

)4(2

x

x

a

bx

This may look complicated, but is VERY helpful…from this we can see that the VS for any general quadratic is a.


For example: Graph the function y = –2x2 + 4x + 6.To complete the square on thistakes time. But…


24

1

a

bx

a

bcy

a

Since the VS = –2 we can graph from the vertex (1, 8):Over 1 down 2Over 2 down 8Over 3 down 18

Shortcut within a shortcut

Instead of memorizing the formula for the y-coordinate of the vertex (VT):

we can calculate it using the general form of the equation and thex-coordinate of the vertex (HT):

For example: What is the maximum value of the function y = –2x2 + 4x + 6?

a

bx

2

8

642

6)1(4)1(2 2

y

y

y

a

bcy

4

2

1

)2(2

)4(2

x

x

a

bx

Max value is y = 8.

General form shortcuts: Practice

4

7

2

1

4

1a) 2 xxy 42243b) 2 xx=y

For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.

General form shortcuts: Practice Solutions4

7

2

1

4

1a) 2 xxy

24

84

7

4

2

4

14

7

2

1

4

14

7)1(

2

1)1(

4

1 2

y

y

y

y

y

12121

41

2

21

2

x

a

bx

vertex (1, –2), VS = 1/4

axis of symmetry: x = 1range: {y ≥ 2, y R} 2124

:form tionaltransforma

xy

General form shortcuts: PracticeSolutions42243b) 2 xx=y

6

429648

42)4(24)4(3 2

y

y

y

46

24

32

242

x

a

bx

vertex (–4, –6), VS = 3

axis of symmetry: x = –4range: {y ≥ –6, y R} 246

3

1

form tionaltransforma

x=y

Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2.

Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation.

A quadratic equation in general form looks like this:ax2 + bx + c = 0 where a ≠ 0.

Solving a quadratic equation for x is also calledfinding its roots or finding its zeros.

Solving for x given y using general form

Example 1:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = –12.

Solution:−12 = x2 – 2x – 15 0 = x2 – 2x – 3

Although the quadratic function had coefficients a = 1, b = –2, and c = –15,subbing in −12 for y gives the quadratic equation with coefficientsa = 1, b = –2, and c = –3

NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example.


NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example.

Example 2:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = 0.

Solution:0 = x2 – 2x – 15

Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15.

While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.


Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form…

So lets try that!

Example 3:Find the x-intercepts of the quadratic function y = x2 – 2x – 15

Solving for x-intercepts using general form

Solution:0 = x2 – 2x – 15…but now what?

2)1(16)0( x2)1(16 x

)1(16 x

14 x3

14

x

x

5

14

x

x

Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.

STEP 1: Write in transformational form by completing the square

STEP 2: Set f(x) = y = 0

STEP 3: Simplify left-hand side

STEP 4: Take the squarroot of both sides

STEP 5: Isolate x in both equations

2

2

2

2

)1(16

12115

215

152

xy

xxy

xxy

xxy

Solving for x-intercepts using general form

Find the roots of the following quadratics:

1144

1c) 2 xxy

442d) 2 xxy

xxy 63a) 2

6416b) 2 xxy

Solving for x-intercepts using general formPractice

A shortcut…eventually

But this is a lot of work….

Let’s find a shortcut to finding x-intercepts by doing this just ONE more time, but with the general equation.

cbxaxy 21522 xxy




STEP 4: Take the square-root of both sides


A shortcut…eventuallycbxaxy 2

1522 xxy2)1(16 xy

22

22

1

a

bx

a

b

a

cy

a22

22)0(

1

a

bx

a

b

a

c

a

2)1(16)0( x

2

2

2

24

a

bx

a

b

a

c

2

2

2

244

)(4

a

bx

a

b

aa

ca

2)1(16 x

2

2

2

2 244

4

a

bx

a

b

a

ac

2

2

2

24

4

a

bx

a

acb




A shortcut…eventually2)1(16 x

STEP 4: Take the square-root of both sides


2

2

2

24

4

a

bx

a

acb

)1(16 x

14 x

a

bx

a

acb

24

42

2

a

bx

a

acb

22

42

xa

acb

a

b

2

4

2

2

a

acbbx

2

42

3

14

x

x

5

14

x

x

This is the shortcut, the Quadratic Root Formula!

Back to our example:Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15

Solution:Instead of those 5 steps, let’s apply the Quadratic Root Formula.

2

822

642

2

6042

12

151422

2

4

2

2

x

x

x

x

a

acbbx

52

82

x

x

32

82

x

x

Solving for x-intercepts using general formThe short-cut

The quadratic root formula can be used to find x-values other than the x-intercepts


For the function f(x) = x2 – 2x – 15, find the values of x when y = −12:

2

422

162

2

1242

12

31422

2

4

2

2

x

x

x

x

a

acbbx

32

42

x

x

12

42

x

x

Solution:–12 = x2 – 2x – 150 = x2 – 2x – 3.

Now use the quadratic root formua witha = 1, b = –2 and c = –3

Find the roots of the following quadratics:

1144

1c) 2 xxy

442d) 2 xxy

xxy 63a) 2

6416b) 2 xxy

Solving for x-intercepts using general formPractice

Finding roots from General Form: Practice Solutions

0 let intercept,- for

63a) 2

yx

xxy


6416b) 2

yx

xxy

2,06

666

366

)3(2

)0)(3(4)6()6(

6302

2

xx

x

x

x

xx

82

016

)1(2

)64)(1(4)16()16(

641602

2

x

x

x

xx



1144

1c) 2

yx

xxy


442d) 2

yx

xxy

528

21

11164

41

2

1141

4)4()4(

1144

10

2

2

x

x

x

xx

11

4

144

4

164

4

32164

)2(2

)4)(2(4)4()4(

44202

2

x

x

x

x

x

xx

Find the x-intercepts of the following quadratic functions:


0,2a) xx

8b) x

528c) x

11d) x

Answers:

1144

1c) 2 xxy

442d) 2 xxy

xxy 63a) 2

6416b) 2 xxy

Yet another method of finding roots: Factoring

As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster!

The zero property:If (s)(t) = 0 then s = 0, or t = 0, or both.(This only works when the product is 0.)

So if we can get the quadratic function in the form:

y = (x – r1)(x – r2), (called factored form)

then when y = 0 we know that:(x – r1) = 0, or (x – r2) = 0, or both.

In other words, x = r1 or x = r2 or both.

Answer when the numbers are added

Factoring to find the roots

We need to factor x2 – 2x – 15.We need 2 numbers whose product is –15 and whose sum is –2

Answer when the numbers are

multiplied

___ × ___ = –15___ + ___ = – 2

-5-5

33

Ex. Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15

(There is never more than one pair of numbers that work!)

So the factored form off(x) = x2 – 2x – 15 isf(x) = (x – 5)(x + 3)

So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3

So the x-intercepts of the graph are (5, 0) and (–3, 0)

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

Ex. Find the roots ofy = x2 + 10x – 24by factoring.

24102 xxy ___ x ___ = -24

___ + ____ = 10)2)(12( xxy

)12(0 x or )2(0 x12x 2x

By the way:(0) = x2 + 10x – 24 24 = x2 + 10x24 + 25 = x2 + 10x + 2549 = (x + 5)2

±7 = x + 5x = –5 + 7 or x = –5 – 7x = 2 or x = –12

12or22

24or

2

42

14102

19610

)1(2

)24)(1(41010 2

xx

xx

x

x

x

)2)(12(0 xx

12

12

-2

-2

Practice1. 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 and 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)

the powers of general form. probe below are 5 different ways of representing a quadratic...

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