A function f defined by f(x) = x² + px + q is such that f(3) = 6 and f¹(3) = 0. Find the value of q.

q = 15

Step-by-step explanation:

Given

f(x) = x² + px + q , then

f(3) = 3² + 3p + q = 6 , that is

9 + 3p + q = 6 ( subtract 9 from both sides )

3p + q = - 3 → (1)

f'(x) = 2x + p , then

f'(3) = 2(3) + p = 0, that is

6 + p = 0 ( subtract 6 from both sides )

p = - 6

Substitute p = - 6 into (1)

3(- 6) + q = - 3

- 18 + q = - 3 ( add 18 to both sides )

q = 15

dont think so

step-by-step explanation:

(-5,-28)

i graphed to on desmos.