Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 8.1
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Exercise 8 Page 374

Practice makes perfect
a

Let's calculate the corresponding y-values for the given x-values. Notice that we will only be substituting integers for x.

c|c|c x & (x-1)^2(x+1) & y [0.2em] [-0.6em] - 2 & ( - 2-1)^2( - 2+1) & - 9 [0.4em] [-0.6em] - 1 & ( - 1-1)^2( - 1+1) & 0 [0.4em] [-0.6em] 0 & ( 0-1)^2( 0+1) & 1 [0.4em] [-0.6em] 1 & ( 1-1)^2( 1+1) & 0 [0.4em] [-0.6em] 2 & ( 2-1)^2( 2+1) & 3Now we can make the value table.

x y
- 2 - 9
- 1 0
0 1
1 0
2 3

Finally, we will place the points in a coordinate plane and draw the graph.

b

Let's calculate the corresponding y-values for the given x-values. Notice that we will only be substituting integers for x.

c|c|c x & (x-1)^2(x+1)^2 & y [0.2em] [-0.6em] - 2 & ( - 2-1)^2( - 2+1)^2 & 9 [0.4em] [-0.6em] - 1 & ( - 1-1)^2( - 1+1)^2 & 0 [0.4em] [-0.6em] 0 & ( 0-1)^2( 0+1)^2 & 1 [0.4em] [-0.6em] 1 & ( 1-1)^2( 1+1)^2 & 0 [0.4em] [-0.6em] 2 & ( 2-1)^2( 2+1)^2 & 9Now we can make the value table.

x y
- 2 9
- 1 0
0 1
1 0
2 9

Finally, we will place the points in a coordinate plane and draw the graph.

c

Let's calculate the corresponding y-values for the given x-values. Notice that we will only be substituting integers for x.

c|c|c x & x^3-4x& y [0.2em] [-0.6em] - 2 & ( - 2)^3-4( - 2) & 0 [0.4em] [-0.6em] - 1 & ( - 1)^3-4( - 1) & 3 [0.4em] [-0.6em] 0 & ( - 1)^3-4( 0) & 0 [0.4em] [-0.6em] 1 & 1^3-4( 1) & - 3 [0.4em] [-0.6em] 2 & 2^3-4( 2) & 0 [0.4em]Now we can make the value table.

x y
- 2 0
- 1 3
0 0
1 - 3
2 0

Finally, we will place the points in a coordinate plane and draw the graph.

d

A parent function is the most basic of functions within the same family of functions. Let's list a few parent functions for some family of functions.

Family of functions Parent function
Linear y=x^()
Quadratic y=x^2
Cubic y=x^3

The function from Part C is a cubic function. Therefore, we can immediately identify its parent function. c. y=x^3-4x → y=x^3 In order to determine the parent function for the remaining equations, we should first rewrite them so that we may identify their family.

y=(x-1)^2(x+1)
y=(x-1)(x-1)(x+1)
â–¼
Simplify right-hand side
y=(x-1)(x^2-1^2)
y=(x-1)(x^2-1)
y=x^3-x-x^2+1
y=x^3-x^2-x+1

The function from Part A is a cubic function. With this information we can identify its parent function. a. y=(x-1)^2(x+1) → y=x^3 Let's also determine the parent functions for the equation from Part B.

y=(x-1)^2(x+1)^2
y=(x-1)(x-1)(x+1)(x+1)
â–¼
Simplify right-hand side
y=(x-1)(x+1)(x-1)(x+1)
y=(x^2-1^2)(x^2-1^2)
y=(x^2-1)(x^2-1)
y=x^4-x^2-x^2+1
y=x^4-2x^2+1

As we can see, the function from Part B is a fourth degree function. With this information, we can identify its parent function. b. y=(x-1)^2(x+1)^2 → y=x^4