Question

How do I find the vertex when it is in my equation is in standard form?

Answer 1

-b/2a = x value

Plug the result into the original equation to solve for y.

Answer 2

Thanksss

Answer 3

You're welcome. Good luck!

Answer 4

Would you also happen to know about how to find the axis of symmetry

Answer 5

-b/2a is also the axis of symmetry.

Answer 6

Alternatively you can convert from standard to vertex form by completing the square:
\[\begin{align*}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\
&=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\
&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)\\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}
\end{align*}\]
which is a parabola with a vertex at
\[\left(-\frac{b}{2a},~\frac{4ac-b^2}{4a}\right)\]