The Minimum Value of a Quadratic Function: A Question of Symmetry

[Monoxdifly]

A quadratic function $$f(x)=x^2+2px+p$$ has the minimum value of –p with $$p\neq0$$. If the curve's symmetrical axis is x = a, then a – f(a) = ...
A. –6
B. –4
C. 4
D. 6
E. 8

Because the curve's symmetrical axis is x = a, then:
$$-\frac{2p}{2(1)}=a$$
–p = a

a – f(a) = –p + (–p) = 0

I got zero. Is there anything I did wrong?

 

[skeeter]

$a – f(a) = –p + (–p) = 0

$a - f(a) = {\color{red}-p - (-p)} = -p + p = 0$

agree with zero ... maybe recheck the original problem statement?

 

[Opalg]

If the problem is indeed asking for $a - f(a)$ then the answer is zero. However ...

The minimum value occurs on the axis of symmetry. Therefore the minimum value is $f(a) = f(-p) = (-p)^2 + 2p(-p) + p = p-p^2$. But you are told that the minimum value is $-p$. Therefore $p-p^2 = -p$, and since $p\ne0$ it follows that $p=2$. Hence $a = -2$, and $f(a)$ is also $-2$. Therefore $a-f(a) = 0$, as we already knew. BUT, if the quetion was actually asking for $a\;{\color{red}+}\,f(a)$ then that would be $-2-2 = -4$, which has the advantage of being one of the multiple choices.

So I agree with skeeter that you should recheck the original problem statement, and in particular look again at whether it is actually asking for $a+f(a)$.

 

[Monoxdifly]

I checked the problem and it said a – f(a), so probably the writer didn't press the Shift button correctly when he/she intended to type "+".