## Algebra and Trigonometry 10th Edition

The vertex is $(-\frac{b}{2a},\frac{-b^2+4ac}{4a})=(-\frac{b}{2a},f(-\frac{b}{2a}))$
$f(x)=ax^2+bx+c$ $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ $f(-\frac{b}{2a})=\frac{b^2}{4a}-\frac{b^2}{2a}+c=-\frac{b^2}{4a}+\frac{4ac}{4a}=\frac{-b^2+4ac}{4a}$ Standard form: $f(x)=a(x-k)^2+h$, in which $(k,h)$ is the vertex: $f(x)=ax^2+bx+c$ $f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$ $f(x)=a[x^2+2(\frac{b}{2a})x+(\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{4ac}{4a^2}]$ $f(x)=a[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}]$ $f(x)=a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}$ So the vertex is $(-\frac{b}{2a},\frac{-b^2+4ac}{4a})=(-\frac{b}{2a},f(-\frac{b}{2a}))$