# How do you find the vertex of y=-x^2+4x+12?

Mar 15, 2018

Factor the equation such that your get $y = - {\left(x - 2\right)}^{2} + 16$ and your vertex appears at (2,16)

#### Explanation:

when you factor a quadratic to find the vertex, it takes the form of:

$y = a {\left(x - h\right)}^{2} + k$, where your vertex is $\left(h , k\right)$

Note that $a$ is the same $a$ as the original quadratic equation. For this equation:

$y = \left(- 1\right) {\left(x - h\right)}^{2} + k \Rightarrow y = \left(- 1\right) \left({x}^{2} - 2 x h + {h}^{2}\right) + k$

$y = - {x}^{2} + 2 x h - {h}^{2} + k$

If we rearrange the above equation to satisfy the other two terms in the original quadratic:

$4 x = 2 x h$
and
$12 = k - {h}^{2}$

Solving the first equation:
$4 = 2 h \Rightarrow \textcolor{red}{h = 2}$

...and now the second:
$12 = k - {\left(2\right)}^{2} \Rightarrow 12 = k - 4 \Rightarrow \textcolor{b l u e}{k = 16}$

Now we have our factors for the vertex form of the quadratic, and we have our vertex!

$y = - {\left(x - 2\right)}^{2} + 16$
$\text{Vertex} = \left(2 , 16\right)$

There are other, quicker ways to compute this too:

$h = \frac{- b}{2 a}$, then plug $h$ into the quadratic to find $k$

If you use calculus, you can take the derivative of the quadratic to find $h$, the vertex is where the slope is 0 so we can find h by plugging in where the derivative goes to 0... although this is literally the same as using $h = \frac{- b}{2 a}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 a x + b$

$0 = 2 a h + b \Rightarrow h = \frac{- b}{2 a}$