Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.5 - Quadratic Functions and Their Graphs - Exercise Set - Page 811: 52

Answer

Please see the graph. The red line is the original function, and the blue line is $y=f(x)-2$.

Work Step by Step

The graph of $f(x)$ has the points as follows: $(-5, -2)$, $(-4, -1)$, $(-3, 0)$, $(-2, 1)$, $(-1, .5)$, $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(5,5)$ For the function $y=f(x)-2$, we decrease the y-value of all points of $f(x)$ by two units. This means our new function has the points $(-5, -4)$, $(-4, -3)$, $(-3, 2)$, $(-2, -1)$, $(-1, -1.5)$, $(0,2)$, $(1,3)$, $(2,4)$, $(3,5)$, $(4,6)$, $(5,7)$
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