Question

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!)

Answer 1

`a = -6` and `b=18` making `f(x) = x^2 -6x +18`

Explanation:

We have:

`f(x) = x^2 + ax + b`

We know that `y=f(x)` passes through `(3,9)`, thus:

`9 = 3^2+3a+b=> 3a+b = 0 .... [A]`

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'(x) = 2x+a`

A critical points occurs when:

`f'(3)=0 => 6+a = 0 => a =-6`

Substituting into Eq [A] we get:

`-18 + b =0 => b = 18`

Hence:

`a = -6` and `b=18` making `f(x) = x^2 -6x +18`

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