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 00 51 82 93 8 4 55 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 00 51 82 93 8 4 55 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(VS1

xy

STEP 1:Divide every term by a.

2621

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

6221

2621

4122

2

2

2


2)HT()VT(VS1

xy

Completing the Squarexxy 62

21 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(1121

363221

6221

2621

4122

xy

xxy

xxy

xxy

xxy


2)HT()VT(VS1

xy



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

2

2

2

2

2

2

)3(2221

)3(1121

969221

6221

2621

4122

xy

xy

xxy

xxy

xxy

xxy


2)HT()VT(VS1

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(21

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(VS1

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.

1421 2 xxy

xxy 4121 2

444121 2 xxy

2)2(321

xy

2)2()6(21

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 xy2)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 xxyxxy 622 2

96922 2 xxy2)3(112 xy

2)3(2112

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

3221 2 xxy

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




xxy 2321 2

222 121321

xxy

211321

xy

21421

xy

21821

xy

cbxaxy 2

acx

abxy

a 21

xabx

acy

a 21

22

2

221

abx

abx

ab

acy

a22

221

abx

ab

acy

a22

241

abx

abcy

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

abx

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

ab2

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

241

abx

abcy

a


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

abc4

2


241

abx

abcy

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

yy

y

abcy

vertex (1, 8)

1)2(2)4(

2

x

x

abx

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…


241

abx

abcy

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 shortcutInstead 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?

abx

2

8642

6)1(4)1(2 2

yyy

abcy4

2

1)2(2)4(

2

x

x

abx

Max value is y = 8.

General form shortcuts: Practice

47

21

41a) 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

721

41a) 2 xxy

248

47

42

41

47

21

41

47)1(

21)1(

41 2

y

y

y

y

y

12121

412

21

2

x

abx

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

6429648

42)4(24)4(3 2

yyy

46243224

2

x

abx

vertex (–4, –6), VS = 3

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

31

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

x5

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

xyxxyxxyxxy

Solving for x-intercepts using general form

Find the roots of the following quadratics:

11441c) 2 xxy

442d) 2 xxy

xxy 63a) 2

6416b) 2 xxy

Solving for x-intercepts using general formPractice

A shortcut…eventuallyBut 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

221

abx

ab

acy

a22

22)0(1

abx

ab

ac

a

2)1(16)0( x

2

2

2

24

abx

ab

ac

2

2

2

244)(4

abx

ab

aaca

2)1(16 x

2

2

2

2 2444

abx

ab

aac

2

2

2

244

abx

aacb




A shortcut…eventually2)1(16 x

STEP 4: Take the square-root of both sides


2

2

2

244

abx

aacb

)1(16 x14 x

abx

aacb

2442

2

abx

aacb

2242

xa

acbab

2

42

2

aacbbx

242

314

xx

514

xx

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.

2822

6422

6042

12151422

24

2

2

x

x

x

x

aacbbx

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:

2422

1622

1242

1231422

24

2

2

x

x

x

x

aacbbx

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:

11441c) 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

2,06666366

)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



11441c) 2

yx

xxy


442d) 2

yx

xxy

52821

11164

412

11414)4()4(

114410

2

2

x

x

x

xx

114

1444

1644

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:

11441c) 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

12or2224or

24

21410

219610

)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

Documents

Transcript of The powers of General Form