## Algebra 2 (1st Edition)

The vertex form of the function is $y=(x-(-\displaystyle \frac{9}{2}))^{2}+(-\frac{5}{4}).$ The vertex is $(-\displaystyle \frac{9}{2},-\frac{5}{4})$.
$y=x^{2}+9x+19\qquad$ ...write in form of $x^{2}+bx=c$ (add $-19$ to each side). $y-19=x^{2}+9x\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{9}{2})^{2}=\frac{81}{4}\qquad$ ...complete the square by adding $\displaystyle \frac{81}{4}$ to each side of the expression $y-19+\displaystyle \frac{81}{4}=x^{2}+9x+\frac{81}{4}\qquad$ ... write $x^{2}+9x+\displaystyle \frac{81}{4}$ as a binomial squared. $y+\displaystyle \frac{5}{4}=(x+\frac{9}{2})^{2}\qquad$ ...add $-\displaystyle \frac{5}{4}$ to each side of the expression $y=(x+\displaystyle \frac{9}{2})^{2}-\frac{5}{4}\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$. $y=(x-(-\displaystyle \frac{9}{2}))^{2}+(-\frac{5}{4})$ The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph. Here, $h=-\displaystyle \frac{9}{2},\ k=-\displaystyle \frac{5}{4}$, so the vertex is $(-\displaystyle \frac{9}{2},-\frac{5}{4})$