Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set: 13

Answer

$\text{$y-$intercept: } -12$

Work Step by Step

Setting $x=0$ and then solving for $y$, then the $y-$intercept of the given equation, $ y=-x^2+8x-12 ,$ is \begin{array}{l}\require{cancel} y=-(0)^2+8(0)-12 \\\\ y=-0+0-12 \\\\ y=-12 .\end{array} Hence, the $y-$intercept is \begin{array}{l}\require{cancel} \text{$y-$intercept: } -12 .\end{array}
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