Find a second degree polynomial f(x) (of the form ax2+bx−4) that has a local extrema at (−1/2,−9/2).

Answer 1

let f(x) = ax^2 + bx - 4

if the vertex is (-1/2, -9/2) then its equation could be
y = a(x + 1/2)^2 - 9/2
= a(x^2 + x + 1/4) - 9/2
= ax^2 + ax + (1/4)a - 9/2

matching them up:
ax^2 <---> ax^2 , ax <---> bx , (1/4)a - 9/2 <---> -4

starting with (1/4)a - 9/2 <---> -4

a - 18 = -16
a = 2
ax <---> bx -----> b=2

so your equation is f(x) = 2x^2 + 2x - 4

check by completing the square:
y = 2x^2 + 2x - 4
= 2(x^2 + x + 1/4 - 1/4) - 4
= 2( (x+1/2)^2 - 1/2 - 4
= 2(x+1/2)^2 - 9/2 , so the vertex is (-1/2, 9/2)
yup, I am right