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Chapter 3 Summary

KeY TeRmS

• standard form of a quadratic function (3 .1)

• factored form of a quadratic function (3 .1)

• vertex form of a quadratic function (3 .1)

• concavity of a parabola (3 .1)• reference points (3 .2)• transformation (3 .2)• rigid motion (3 .2)• argument of a function (3 .2)• translation (3 .2)• vertical dilation (3 .3)• vertical stretching (3 .3)

• vertical compression (3 .3)• reflection (3 .3)• line of reflection (3 .3)• horizontal dilation (3 .4)• horizontal stretching (3 .4)• horizontal compression (3 .4)• the imaginary number i (3 .6)• principal square root of a

negative number (3 .6)• set of imaginary numbers (3 .6)• pure imaginary number (3 .6)• set of complex numbers (3 .6)• real part of a complex

number (3 .6)

• imaginary part of a complex number (3 .6)

• complex conjugates (3 .6)• monomial (3 .6)• binomial (3 .6)• trinomial (3 .6)• imaginary roots (3 .7)• discriminant (3 .7)• imaginary zeros (3 .7)• Fundamental Theorem of

Algebra (3 .7)• double root (3 .7)

Using Characteristics of a Quadratic Function to Describe Its GraphThe graphs of quadratic functions can be described using key characteristics:

• x-intercept(s)

• y-intercept

• vertex

• axis of symmetry

• concave up or down

The y-intercept (c) and whether the parabola is concave up (a . 0) or down (a , 0) can be determined when the quadratic is in standard form, f(x) 5 ax2 1 bx 1 c .

The x-intercepts (r1, 0), (r2, 0), and whether the parabola is concave up (a . 0) or down (a , 0) can be determined when the quadratic is in factored form, f(x) 5 a(x 2 r1)(x 2 r2) .

The vertex (h, k), whether the parabola is concave up (a . 0) or down (a , 0), and the axis of symmetry (x 5 h) can be determined when the quadratic is in vertex form, f(x) 5 a(x 2 h)2 1 k .

The example shows that using the key characteristics of a graph can also determine the function represented by the graph .

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Example

Analyze the graph . Then, circle the function(s) which could model the graph . Describe the reasoning you used to either eliminate or choose each function .

x

y

f1(x) 5 2(x 2 3)(x 1 4)

The function f1 can be eliminated because the a-value is positive .

f2(x) 5 1 __ 2 (x 1 3)(x 2 4)

The function f2 can be eliminated because the a-value is positive .

f3(x) 5 2 1 __ 2 (x 1 3)(x 2 4)

The function f3 is a possible function because it has a negative a-value and one positive and one negative x-intercept .

f4(x) 5 2 1 __ 2 (x 2 3)(x 2 4)

The function f4 can be eliminated because it has two positive x-intercepts .

Determining the Best Form in Which to Write a Quadratic FunctionGiven the roots or x-intercepts and an additional point, a quadratic function can be written in factored form: f(x) 5 a(x 2 r1)(x 2 r2) .

Given a minimum or maximum point or vertex and an additional point, a quadratic function can be written in vertex form: f(x) 5 a(x 2 h)2 1 k .

Given a y-intercept and two additional points, a quadratic function can be written in standard form: f(x) 5 ax2 1 bx 1 c .

Example

vertex (2, 25) and point (3, 1): vertex form

points (2, 0), (6, 5), and (9, 0): factored form

y-intercept (0, 22) and points (24, 3), (5, 2): standard form

Writing a Quadratic Function in Standard FormWhen a quadratic function is in vertex or factored form, the function can be rewritten in standard form by multiplying the factors and combining like terms .

Example

f(x) 5 (x 2 2)(x 1 8) f(x) 5 22(x 2 3)2 1 6

f(x) 5 (x 2 2)(x 1 8) f(x) 5 22(x 2 3)2 1 6

5 x2 1 8x 2 2x 2 16 5 22(x2 2 6x 1 9) 1 6

5 x2 1 6x 2 16 5 22x2 1 12x 2 18 1 6

5 22x2 1 12x 2 12

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Writing a Quadratic Function to Represent a SituationDetermine the important information from the problem situation, including a possible vertex or x-intercepts . Substitute the information into either the vertex or factored form of a quadratic function and solve for a . Write the complete function .

Example

Hector launches a rocket from a platform 5 feet above the ground . The rocket reaches a maximum height of 50 feet at a distance of 75 feet . Write a function to represent the rocket’s height in terms of its distance .

h(d) 5 a(d 2 50)2 1 75

5 5 a(0 2 50)2 1 75

5 5 2500a 1 75

270 5 2500a

27 ____ 250

5 a

h(d) 5 2 7 ____ 250 (d 2 50)2 1 75

Identifying the Reference Points of a Quadratic FunctionReference points are a set of key points that help identify a basic function .

ReferencePointsofBasicQuadraticFunction

P (0, 0)

Q (1, 1)

R (2, 4)

If the vertex of a quadratic function and two points to the right of that vertex are known, the axis of symmetry can be used to draw the other half of the parabola . These key points are reference points for the quadratic function family .

x

2222 P(0, 0)24 4

y

2

4

24

Q(1, 1)

R(2, 4)

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Understanding the effect of the C-Value and the D-Value in the Transformational Function Form g(x) 5 Af(B(x 2 C)) 1 DA translation shifts an entire graph the same distance and direction . When f(x) is transformed to g(x) 5 Af(B(x 2 C)) 1 D, the C-value shifts the graph horizontally and the D-value shifts the graph vertically . When C , 0, the graph shifts left C units, and when C . 0, the graph shifts to the right C units . When D , 0, the graph shifts down D units, and when D . 0, the graph shifts up D units .

Example

f(x) 5 x2

m(x) 5 (x 1 3) 1 1

C 5 23 and D 5 1

The original function f(x) will translate to the left 3 units and translate up 1 unit .

ReferencePointsonf(x)

→ ApplytheTransformations

CorrespondingPointsonm(x)

(0, 0) → (0 2 3, 0 1 1) (23, 1)

(1, 1) → (1 2 3, 1 1 1) (22, 2)

(2, 4) → (2 2 3, 4 1 1) (21, 5)

x

2

4

2242024 22 8628 26

y

24

26

28

6

8m(x)

Writing the Function of a Transformed GraphIdentify the location of the new vertex . Use the x- and y-coordinates of the new vertex as the C- and D-values . Write the transformed function in terms of the original function in the form g(x) 5 f(x 2 C) 1 D .

Example

Given y 5 f(x)

g(x) 5 f(x 2 2) 2 4

3.2

3.2

x

1

2

2121022 21 43 5

f(x)g(x)

23

y

22

23

24

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Chapter 3 Summary 283

Understanding the effect of the A-Value in the Transformational Function Form g(x) 5 Af(B(x 2 C)) 1 DA translation shifts an entire graph the same distance and direction . When f(x) is transformed to g(x) 5 Af(B(x 2 C)) 1 D, the A-value stretches or compresses the graph vertically . The C-value shifts the graph horizontally and the D-value shifts the graph vertically . When |A| $ 1, the graph stretches away from the x-axis, and when 0 , |A| , 1, the graph compresses towards the x-axis . When A , 0, the graph is reflected across the x-axis or across the horizontal line y 5 D . Any point on the transformed graph can be represented in coordinate notation as

(x, y) → ( 1 __ B

x 1 C, Ay 1 D ) .Example

f(x) 5 x2

b(x) 5 24(f(x 1 1) 2 3

The C-value is 21 and the D- value is 23 so the vertex will be translated 1 unit to the left and 3 units down to (21, 23) .

The A-value is 24, so the graph will have a vertical stretch by a factor of 4 and will be reflected across the line y 5 23 .

(x, y) 5 (x 2 1, 24y 2 3)

Writing a Quadratic Function to Represent a GraphGiven y 5 f(x), the coordinate notation represented in the transformational function y 5 Af(x 2 C) 1 D is (x, y) becomes (x 1 C, Ay 1 D) . Using this coordinate notation and reference points, you can work backwards from the graph to write the function in vertex form .

Example

x

2

4

2242024 22 8628 26

y

24

26

28

6

8

f(x) 5 1 __ 3 (x 2 3)2 2 4

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Graphing Horizontal Dilations of Quadratic FunctionsConsidering the transformational function g(x) 5 Af(B(x 2 C)) 1 D, B determines the horizontal dilation .

• Horizontal stretching is the stretching of a graph away from the y-axis and happens when 0 , |B| , 1 .

• Horizontal compression is the squeezing of a graph towards the y-axis and happens when |B| . 1 .

The factor of horizontal stretch or compression is the reciprocal of the B-value . When B , 0, the dilation will also include a reflection across the y-axis, but because a parabola is symmetric, it will look identical to the function when B . 0 .

Example

p(x) 5 f(2x)

ReferencePointsonf(x)

→ ApplytheTransformations

CorrespondingPointsonp(x)

(0, 0) → ( 1 __ 2

? 0, 0 ) (0, 0)

(1, 1) → ( 1 __ 2

? 1, 1 ) (0 .5, 1)

(2, 4) → ( 1 __ 2

? 2, 4 ) (1, 4)

(3, 9) → ( 1 __ 2

? 3, 9 ) (1 .5, 9)

x

2

4

21022 21 4324 23

y

6

8

10

12

14

16

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Identifying multiple Transformations Given Quadratic FunctionsAnalyzing the transformations of y 5 f(x), it is possible to graph a function from information given by the form of the equation .

FunctionForm EquationInformation DescriptionofTransformationofGraph

y 5 Af (B(x 2 C)) 1 D

D . 0 vertical shift up D units

D , 0 vertical shift down D units

C . 0 horizontal shift right C units

C , 0 horizontal shift left C units

|A| . 1 vertical stretch by a factor of A units

0 , |A| , 1 vertical compression by a factor of A units

A , 0 reflection across the line y 5 k

|B| . 1 horizontal compression by a factor of 1 ___ |B|

0 , |B| , 1 horizontal stretch by a factor of 1 ___ |B|

B , 0 reflection across the y-axis

Exampleh(x) 5 23(x 1 7)2 2 8

The A value is 23 so the graph will have a vertical stretch by a factor of 3 and be reflected over the x-axis . The C-value is 27 and the D-value is 28, so the vertex will be shifted 7 units to the left and 8 units down to (27, 28) . The A-value is 23 so the graph will have a vertical stretch by a factor of 3 and be reflected across the line y 5 D, or y 5 28

Example

Given the graph of y 5 f(x), the graph of g(x) 5 2f ( 1 __ 2 (x 2 4) ) 2 5 is sketched .

x

2

4

2242024 22 86 10

f(x)

g(x)

26

y

24

26

28

6

8

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Writing Quadratic Functions in Terms of a Given a GraphDetermine by how much and in what direction the reference points of the graph have moved and if there has been a reflection, horizontal dilation, or vertical dilation . Use coordinate

notation ( 1 __ B

x 1 C, Ay 1 D ) to help identify A, B, C, and D values . Write the transformed

function in terms of the original function in the form g(x) 5 Af(B(x 2 C)) 1 D .

Example

The function of g(x) in terms of f(x) is g(x) 5 f(2x) 2 5, or g(x) 5 f(x 2 12) 2 5 .

Deriving a Quadratic equation Given Certain Information and Reference PointsTo write a unique quadratic function, use the reference points of the basic quadratic function and consider the vertical distance between each point . If given the x-intercepts and another point on the parabola, a function can be written in factored form or if given the vertex and another point on the parabola, a function can be written in vertex form; but in both cases the a-value must also be determined . To do this, first determine the axis of symmetry as directly in the middle of the two x-intercepts or at the location of the vertex . Plot the axis of symmetry and points on a coordinate plane and assign each point a letter based on its horizontal distance from the line of symmetry . The vertical distance between A9 and B9 will be the vertical distance between A and B of the basic function times a .

Example

x-intercepts and 1 point: (23, 0), (5, 0), (4, 14)

r1 1 r2 ______

2 5 23 1 5 _______

2 5 2 __

2 5 1 so the axis of

symmetry is x 5 1 . Point (5, 0) is point D9 because it is 4 units from the axis of symmetry . Point (4, 14) is point C9 because it is 3 units from the axis of symmetry . The range between the C and D point on the basic function is 7 . The range between point D9 and point C9 is 7 3 (22) 5 214, therefore the a-value must be 22 .

f(x) 5 22(x 1 3) (x 2 5)

3.4

x

2

4

2242024 22 86

f(x)

g(x)

2628

y

24

26

28

6

8

3.5

x

2

4

21022 21 4

14

(5, 0) D9(23, 0)

C9(4, 14)

3 units

4 units324 23

y

6

8

10

12

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Chapter 3 Summary 287

Deriving a Quadratic equation Given Three Points Using a Graphing Calculator

You can use a graphing calculator to determine a

quadratic regression equation given three points

on the parabola.

Step 1: Diagnostics must be turned on so that all

needed data is displayed. Press 2nd CATALOG

to display the catalog. Scroll to

DiagnosticOn and press ENTER. Then press

ENTER again. The calculator should display

the word Done.

Step 2: Press STAT and then press ENTER to select

1:Edit. In the L1 column, enter the x-values

by typing each value followed by ENTER. Use

the right arrow key to move to the L2 column.

Enter the y-values.

Step 3: Press STAT and use the right arrow key to

show the CALC menu. Choose 5:QuadReg. Press

Enter. The values for a, b, and c will be

displayed.

Step 4: To have the calculator graph the exact

equation for you, press Y=, VARS,

5:Statistics, scroll right to EQ, press

1:RegEQ, GRAPH.

Example

(28, 1), (21, 13 .25), (6, 248)

f(x) 5 20 .75x2 2 5x 1 9

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Deriving a Quadratic equation Given Three Points Using a System of equationsTo use a system of equations, first create a quadratic equation in the form y 5 ax2 1 bx 1 c for each of the points given . Then, use elimination and substitution to solve for a, b, and c .

Example

(21, 211), (2, 25), (4, 241)

Equation 1: 211 5 a(21)2 1 b(21) 1 c Equation 2: 25 5 a(2)2 1 b(2) 1 c

211 5 a 2 b 1 c 25 5 4a 1 2b 1 c

Equation 3: 241 5 a(4)2 1 b(4) 1 c

241 5 16a 1 4b 1 c

Subtract Equation 1 from Equation 2 and solve in terms of a:

25 5 4a 1 2b 1 c2(211 5 a 2 b 1 c)

6 5 3a 1 3b2 2 b 5 a

Subtract Equation 2 from Equation 3:

241 5 16a 1 4b 1 c2(25 5 4a 1 2b 1 c)

236 5 12a 1 2b218 5 6a 1 b

Substitute the value for a into this equation:

218 5 6(2 2 b) 1 b218 5 12 2 5b230 5 25b

b 5 6

Substitute the value of b into the equation for the value of a:

a 5 2 2 6a 5 24

Substitute the values of a and b into Equation 1:

211 5 24 2 6 1 c211 5 210 1 c

21 5 c

Substitute the values for a, b, and c into a quadratic equation in standard form:

f(x) 5 24x2 1 6x 2 1

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Chapter 3 Summary 289

Calculating Powers of iThe imaginary number i is a number such that i2 5 21 . Because no real number exists such that its square is equal to a negative number, the number i is not a part of the real number system . The values of in repeat after every four powers of i, where i 5 √

___ 21 , i2 5 21, i3 5 2 √

___ 21 ,

and i4 5 1 .

Example

i25 5 (i4)6 (i1)

5 (1)6 ( √____

21 )

5 √____

21

Rewriting expressions with Negative Roots Using iExpressions with negative roots can be rewritten . For any positive real number n, the principal square root of a negative number, 2n, is defined by √

___ 2n 5 i √

__ n .

Example

√_____

263 1 √_____

224 5 i √___

63 1 i √___

24

5 i √_____

(9)(7) 1 i √_____

(6)(4)

5 3 √__

7 i 1 2 √__

6 i

Adding, Subtracting, and multiplying on the Set of Complex Numbers The set of complex numbers is the set of all numbers written in the form a 1 bi, where a and b are real numbers . The term a is called the real part of a complex number, and the term bi is called the imaginary part of a complex number . When operating with complex numbers involving i, combine like terms by treating i as a variable (even though it’s a constant) . Complex conjugates are pairs of numbers of the form a 1 bi and a 2 bi . The product of a pair of complex conjugates is always a real number in the form a2 1 b2 .

Example

4x 1 (x 1 3i )(x 2 3i ) 2 5x 1 7

4x 1 (x 1 3i )(x 2 3i ) 2 5x 1 7 5 4x 1 x2 2 9i2 2 5x 1 7

5 x2 1 (4x 2 5x) 1 (7 2 9(21))

5 x2 2 x 1 16

3.6

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Identifying Complex Polynomials A polynomial is a mathematical expression involving the sum of powers in one or more variables multiplied by coefficients . The definition of a polynomial can be extended to include imaginary numbers . Some polynomials have special names, according to the number of terms they have . A polynomial with one term is called a monomial . A polynomial with two terms is called a binomial . A polynomial with three terms is called a trinomial . Combine like terms to name the polynomial .

Example

5x 2 3xi 1 x2 2 3i 1 9

The expression is a trinomial because it can be rewritten as x2 1 (5 2 3i )x 1 (9 2 3i ), which shows one x2 term, one x term, and one constant term .

Adding, Subtracting, and multiplying Complex Polynomials Some polynomial expressions involving i can be simplified using methods similar to those used to operate with numerical expressions involving i . Whenever possible, multiply the complex conjugates first to get a real number .

Example

(3x 1 7i )(2x 1 5i )

(3x 1 7i )(2x 1 5i ) 5 6x2 1 15xi 1 14xi 1 35i2

5 6x2 1 29xi 1 35(21)

5 6x2 1 29xi 2 35

Rewriting the Quotient of Complex NumbersWhen rewriting the quotient of complex numbers, multiply both the divisor and the dividend by the complex conjugate of the divisor, thus changing the divisor into a real number . The product of a pair of complex conjugates is always a real number in the form a2 1 b2 .

Example

3 2 i ______ 5 1 2i

5 3 2 i ______ 5 1 2i

? 5 2 2i ______ 5 2 2i

5 15 2 6i 2 5i 1 2i2 ________________ 52 1 22

5 15 2 11i 2 2 ____________ 25 1 4

5 13 2 11i ________ 29

5 13 ___ 29 2 11 ___

29 i

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Using the Quadratic Formula to Determine the Zeros of a Function If the graph of f(x) intersects the x-axis 2 times, then the quadratic equation f(x) 5 0 has 2 solutions, or roots . If the graph of f(x) intersects the x-axis 1 time, then the quadratic equation f(x) 5 0 has 1 solution and double roots . If the graph of f(x) does not intersect the x-axis, then the quadratic equation f(x) 5 0 has no solution, or imaginary roots . Use the Quadratic Formula and what you know about imaginary numbers to solve an equation of the form f(x) 5 0 to calculate the roots, or zeros of the function .

Remember the Quadratic Formula is: x 5 2b 6 √_________

b2 2 4ac _______________ 2a

.

Example

g(x) 5 x2 2 7x 1 12

x2 2 7x 1 12 5 0

a 5 1, b 5 27, c 5 12

x 5 2b 6 √_________

b2 2 4ac _______________ 2a

x 5 2(27) 6 √

______________ (27)2 2 4(1)(12) ________________________

2(1)

x 5 7 6 √__

1 _______ 2

x 5 4, x 5 3

Determining Whether a Quadratic Function in Standard Form Has Real or Imaginary Zeros The radicand in the Quadratic Formula, b2 2 4ac, is called the discriminant because it “discriminates” the number and type of roots of a quadratic equation . If the discriminant is greater than or equal to zero, the quadratic equation has two real roots . If the discriminant is negative, the quadratic equation has two imaginary roots . If the discriminant is equal to zero, the quadratic equation appears to have only 1 root, but still has 2 real roots called a double root . The Fundamental Theorem of Algebra states that for any polynomial equation of degree n must have exactly n complex roots or solutions; also, every polynomial function of degree n must have exactly n complex zeros . Zeros of a function f(x) are the values of x for which f(x) 5 0 and are related to the roots of an equation .

Example

f(x) 5 x2 2 3x 1 14

b2 2 4ac 5 (23)2 2 4(1)(14)

5 9 2 56

5 247

The discriminant is negative, so the function has two imaginary zeros .

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Determining Whether a Quadratic Function in Vertex Form Has Real or Imaginary Zeros If the graph of f(x) intersects the x-axis 2 times, then the quadratic equation f(x) 5 0 has 2 solutions, or roots . If the graph of f(x) intersects the x-axis 1 time, then the quadratic equation f(x) 5 0 has 1 solutions and double roots . If the graph of f(x) does not intersect the x-axis, then the quadratic equation f(x) 5 0 has no solution, or imaginary roots . So, by using C and D of vertex form to locate the vertex above or below the x-axis and using A to determine if the parabola is concave up or down, you can tell if the graph intersects the x-axis or not .

Example

f(x) 5 25(x 2 2)2 1 7

Because the vertex (2, 7) is above the x-axis and the parabola is concave down (a , 0), it intersects the x-axis . So, the zeros are real .

Determining Whether a Quadratic Function in Factored Form Has Real or Imaginary Zeros Some functions can be factored over the set of real numbers, while other functions can be factored over the set of imaginary numbers . However, all functions can be factored over the set of complex numbers . If necessary, use the Quadratic Formula to determine the roots and write the function in factored form .

Example

p(x) 5 x2 1 6x 1 18

p(x) 5 [x 2 (23 1 3i )][x 2 (23 2 3i )]

x 5 23 1 3i, x 5 23 2 3i

The function p(x) has two imaginary zeros .

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