Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Absolute Value Functions and Transformations
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Exercise 21 Page 476

Practice makes perfect
a Let's begin by examining the general form of an absolute value function. Therefore, we can understand what h, k, and a stand for.

f(x)=a|x- h|+ k In this form, a tells us how far the graph stretches or shrinks vertically, h and k give an idea about how far the graph shifts horizontally and vertically. To identify h, k, and a, we will rewrite the given equations in the general form. Let's start! f(x)=3|x-2|+2 ⇓ f(x)=3|x- 2|+ 2 The function is already in general form, so the terms can be identified as a=3, h=2, and k=2.

b Since the given function in this part is also in the general form, we will proceed in the same way as we did in Part A.

f(x)=-0.2|x-3|+4 ⇓ f(x)=-0.2|x- 3|+ 4 Therefore, the terms a=-0.2, h=3, and k=4.

c As we can see here, the given function is not in the general form. Let's rewrite the function in general form by changing the signs of h and k.

f(x)=-5|x+6|-1 ⇓ f(x)=-5|x-( -6)|+( -1) Now that the function is in general form, the terms can be identified as a=-5, h=-6, and k=-1.

d To rewrite the given function in general form, we will change the signs of h and k as we did in Part C.

f(x)=-0.5|x+2|-7 ⇓ f(x)=-0.5|x-( -2)|+( -7) From there, the terms are a=-0.5, h=-2, and k=-7.

e When we look at this function, we see that the term h is missing. To rewrite this function in general form, we will represent h as 0.

f(x)=0.8|x|+3 ⇓ f(x)=0.8|x- 0|+ 3 Therefore, the terms are a=0.8, h=0, and k=3.