Question

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

Answer 1

`(3, 12)`

Explanation:

Use `x_(vertex)=(-b)/(2a)`
In this case, `a=-1, b=6`, so `x_(vertex)=3`
Then, the coordinate is `(3, f(3)) = (3, 12)`

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_(vertex)=(x_1 + x_2) / 2`
We also know that:
`x_(1, 2)=(-b+-sqrt(b^2-4ac))/(2a)=(-b+-sqrt(Delta))/(2a)`
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))`
`=(-b)/(2a)`

Voila.