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?

Hint: from your information we know that $y(1) = -4$ and $\frac{\mathrm{d}y(1)}{\mathrm{d}x} = 0$. — molarmass, Jul 04 '15 at 15:38

score 3 · Answer 1 · answered Jul 04 '15 at 15:38

3

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

answered Jul 04 '15 at 15:38

abel

29,170

score 1 · Answer 2 · answered Jul 04 '15 at 15:37

1

Hint:

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

answered Jul 04 '15 at 15:37

YoTengoUnLCD

13,384

1

The OP asks for a solution rather than a hint. In such cases a brief hint as you propose would better serve as a Comment rather than a full Answer. – hardmath Jul 04 '15 at 15:56

score 1 · Answer 3 · answered Jul 04 '15 at 15:39

1

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 $q-p^2/4$.

answered Jul 04 '15 at 15:39

Ángel Mario Gallegos

19,882

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

3 Answers3