# The curve of y=f(x) where f(x) = x^2 + ax + b  has a minimum at (3,9). Find a and b?

## (Question Restore: portions of this question have been edited or deleted!)

Jan 16, 2018

$a = - 6$ and $b = 18$ making $f \left(x\right) = {x}^{2} - 6 x + 18$

#### Explanation:

We have:

$f \left(x\right) = {x}^{2} + a x + b$

We know that $y = f \left(x\right)$ passes through $\left(3 , 9\right)$, thus:

$9 = {3}^{2} + 3 a + b \implies 3 a + b = 0 \ldots . \left[A\right]$

We also require a minimum at this coordinate, (we know that it will have a minimum as we have positive coefficient of ${x}^{2}$ so differentiating wrt $x$ we have:

$f ' \left(x\right) = 2 x + a$

A critical points occurs when:

$f ' \left(3\right) = 0 \implies 6 + a = 0 \implies a = - 6$

Substituting into Eq [A] we get:

$- 18 + b = 0 \implies b = 18$

Hence:

$a = - 6$ and $b = 18$ making $f \left(x\right) = {x}^{2} - 6 x + 18$

Which we confirm graphically:
graph{x^2 -6x +1 [-5, 10, -10, 5]}