Question 476981
(a) f'(x) = 4x+8

(b)  (1,3) is NOT a point on the parabola.  

(c) You don't need calculus to know if there is a max or a min, it's obvious from the coefficient of {{{x^2}}} that the function has an (absolute) min.

BUT if you insist, then f'(x) = 4x+8 = 0 ==> x = -2.

The 2nd derivative is f"(x) = 4 > 0, so by the 2nd derivative test there is (absolute) min at x = -2.  (x = -2 could have been gotten using the good ol' formula {{{-b/(2a)}}}.)