## Algebra 2 (1st Edition)

The vertex form of the function is $f(x)=(x-\displaystyle \frac{3}{2})^{2}+\frac{7}{4}.$ The vertex is $(\displaystyle \frac{3}{2},\frac{7}{4})$.
$f(x)=x^{2}-3x+4\qquad$ ...first, prepare to complete the square. $f(x)+?=(x^{2}-3x+?)+4\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{-3}{2})^{2}=\frac{9}{4}\qquad$ ...complete the square by adding $\displaystyle \frac{9}{4}$ to each side of the expression $f(x)+\displaystyle \frac{9}{4}=x^{2}-3x+\frac{9}{4}+4\qquad$ ... write $x^{2}-3x+\displaystyle \frac{9}{4}$ as a binomial squared. $f(x)+\displaystyle \frac{9}{4}=(x-\frac{3}{2})^{2}+4\qquad$ ...add $-\displaystyle \frac{9}{4}$ to each side of the expression $f(x)+\displaystyle \frac{9}{4}-\frac{9}{4}=(x-\frac{3}{2})^{2}+4-\frac{9}{4}\qquad$ ...simplify. $f(x)=(x-\displaystyle \frac{3}{2})^{2}+\frac{7}{4}$ 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, $h=\displaystyle \frac{3}{2},\ k=\displaystyle \frac{7}{4}$, so the vertex is $(\displaystyle \frac{3}{2},\frac{7}{4})$