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

Aug 3, 2017

Let's look at the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Knowing parabolas are symmetric around the vertex, we can infer that the axis lies on the vertical line defined by the non-variant part of the quadratic formula (a.k.a the one not affected by the $\pm$)

$\text{Non variant part: } \frac{- b}{2 a}$

We know the result of that (after plugging $a = - 1 , b = 4$) will be the x-value of the vertex. To find the y-value, we simply plug in the x-value into the function to see its value.

So, generalising, the vertex of a quadratic function $f \left(x\right)$ is always:

$V \left(\frac{- b}{2 a} , f \left(\frac{- b}{2 a}\right)\right)$

In this case it's $V \left(2 , 16\right)$

graph{-x^2 + 4x +12 [-16.81, 19.23, -0.73, 17.29]}

Aug 3, 2017

$\left(2 , 16\right)$

#### Explanation:

Well, firstly find the axis of symmetry for the parabola. The formula for this axis of symmetry is $- \frac{b}{2 a}$.

Now you may not know what $a$ and $b$ stand for, but this $a {x}^{2} + b x + c$ formula that you have seen this is the general equation of a parabola.

Notice the $a$ in front of the ${x}^{2}$ that is a constant that is your $a$ in the axis of symmetry equation $- \frac{b}{2 a}$. Similarly, the $b$ in front of the $x$ in your general parabola equation is the $b$ in your axis of symmetry equation $- \frac{b}{2 a}$.

So sub in your values for $a$ and $b$ and you will get

$\frac{- 4}{-} 2 = 2$

Thus your axis of symmetry is $x = 2$.

To find your vertex, sub this $2$ into your main equation which was

$y = - {x}^{2} + 4 x + 12$

So

$- {\left(2\right)}^{2} + 4 \cdot 2 + 12 = 16$

Finally, your vertex is at $\left(2 , 16\right)$.