[ASK] a – f(a)

[Monox D. I-Fly]

A quadratic function [tex]f(x)=x^2+2px+p[/tex] has the minimum value of –p with [tex]p\neq0[/tex]. 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:

[tex]-\frac{2p}{2(1)}=a[/tex]

–p = a


a – f(a) = –p + (–p) = 0


I got zero. Is there anything I did wrong?

[romsek]

I get 0 as well.

[Guest]

Complete the square!
[tex]x^2+ 2px+ p= x^2+ 2px+ p^2- p^2+ p= (x+ p)^2- p^2+ p[/tex].

Since a square is never negative the minium value of this is when x= -p and that minimum value is [tex]p- p^2[/tex].

Since we are told that "the curves symmetrical axis is x= a", p must be -a. [tex]f(a)= a^2+ 2(-a)a+ (-a)= -a^2- a= -(a^2+ a)[/tex].

[Guest]

[tex]f(x)= x^{2 }+2px+p[/tex]

[tex]f'(x)= 2x+2p[/tex]

[tex]f'(a)= 0[/tex]

[tex]2a+2p= 0[/tex]

[tex]a+p= 0[/tex]

[tex]f(a)= -p[/tex]

[tex]a-f(a)= a-(-p)=a+p=0[/tex]