Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set - Page 665: 16

$\text{$y-$intercept: } 14$

Setting $x=0$ and then solving for $y$, then the $y-$intercept of the given equation, $ y=x^2+8x+14 ,$ is \begin{array}{l}\require{cancel} y=(0)^2+8(0)+14 \\\\ y=0+0+14 \\\\ y=14 .\end{array} Hence, the $y-$intercept is \begin{array}{l}\require{cancel} \text{$y-$intercept: } 14 .\end{array}