Find $p$ and $q$ for $y(x)=x^2+px+q$.

Question

Find $p$ and $q$ for $y(x)=x^2+px+q$ if the function has minimum equal to $-4$ for $x=1$ Can anyone try to solve this please?

Answer 1

Hint:

$\lambda$ is a maximum or minimum of $f$ implies $f'(\lambda)=0$

Answer 2

with the quadratic having a global min at $(1, -4)$ means we can write $$x^2 + px + q = (x-1)^2 - 4 = x^2 -2x-3.$$

Answer 3

Hint:

Since $$x^2+px+q=\left(x+\frac{p}{2}\right)^2+q-\frac{p^2}{4}\ge 0 +q-\frac{p^2}{4}$$ it follows $y$ reaches its minimum value at $x=-p/2$, and the minimum value is $q-p^2/4$.