Question

How do you convert standard form to vertex form?

Answer 1

This should help!
www.youtube.com/watch?v=rp1iQdCCBnI

Answer 2

By completing the square and rearranging it

Answer 3

Complete the square.

Consider the general case:
\[\text{standard form: }y=ax^2+bx+c~~~~\to~~~~\text{vertex form: }y=a(x-h)^2+k\]
where the vertex is \((h,k)\).
\[\begin{align*}y&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\
&=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-c}{4a^2}\right)\\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-c}{4a}
\end{align*}\]
The vertex would then be
\[(h,k)=\left(-\frac{b}{2a},~\frac{b^2-c}{4a}\right)\]