# What is the vertex of f(x)= -x^2 + 6x + 3?

May 23, 2017

$\left(3 , 12\right)$

#### Explanation:

Use ${x}_{v e r t e x} = \frac{- b}{2 a}$
In this case, $a = - 1 , b = 6$, so ${x}_{v e r t e x} = 3$
Then, the coordinate is $\left(3 , f \left(3\right)\right) = \left(3 , 12\right)$

Derivation of this formula:

We know the vertex's x position is the average of the two solutions. To find the x component of the vertex, we take the average:
${x}_{v e r t e x} = \frac{{x}_{1} + {x}_{2}}{2}$
We also know that:
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$
where $\Delta$ is the discriminate.

So then we can derive that:
x_(vertex)=1/2 ((-b+sqrt(Delta))/(2a) + (-b-sqrt(Delta))/(2a)) =1/2((-b + sqrt(Delta) + -b - sqrt(Delta)) / (2a)) =1/2((-2b)/(2a))
$= \frac{- b}{2 a}$

Voila.