Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set - Page 665: 13

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

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}