Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Mid-Chapter Check Point - Page 629: 15

Answer

Domain of $f$ is (-∞, ∞) Range of $f$ is $[-4,∞)$
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Work Step by Step

$f(x)=(x-3)^2-4$ Expand. $x^2-6x+5$ $a=1$, $b=-6$, $c=5$ Graphing: The coefficient of $x$ is $1$, therefore the parabola that opens upward. Get the vertex: $x=\frac{-b}{2a}$ $x=\frac{-(-6)}{2(1)}$ $x=3$ $f(3)=(3-3)^2-4$ $f(3)=-4$ vertex is at $(3,-4)$ Find the $x$-intercepts by solving $f (x) = 0$. $0=(x-3)^2-4$ $(x-3)^2=4$ Apply the square root property. $x-3=±\sqrt{4}$ $x=2+3$ or $x=-2+3$ $x=5$ or $x=1$ The $x$-intercepts are at points $(5,0)$ and $(1,0)$ Domain: The graph widens and continues to fall at both ends, and thus, include all real numbers Domain of $f$ is (-∞, ∞). Range: The parabola’s vertex, (3, -4), is the lowest point on the graph. Because the y-coordinate of the vertex is $-4$, outputs on the y-axis fall at or above $-4$. Range of $f$ is $[-4,∞)$
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