The equation of the parabola with the sample shape as f(x)=9x^2 but with vertex (-3,2) is?

Question

Possible answers: A. f(x)=9(x+3)^2+2
B. f(x)=(9x+3)^2+2
C. f(x)=9(x-2)^2-3
D. f(x)=9(x-3)^2+2
Please show work. I think it is A but not sure. Thank you

Answer 1

You are correct! Just shift the vertex from (0,0) to (-3,2) and keep the 9 the same (that's what dictates the shape and direction of opening):

{A} f(x) = 9(x+3)² + 2

Answer 2

To calculate a critical point for any curve you must derive the equation and equalize it to zero as follows:

For the parabola equation f(x) = ax^2 + bx + c
f'(x) = 2ax + b = 0
--> x = - b/(2a) and
y = f(x) = a[ -b/(2a)]^2 + b[ -b/2a ] + c
= a[ b^2 / (4a^2) ] - b^2 / (2a) + c
= b^2 / (4a) - b^2 / (2a) + c
= - b^2 / (4a) + c = - ( b^2 - 4ac ) / (4a)

Then The Vertex is:
x = -b / (2a)
y = - ( b^2 - 4ac ) / (4a)

Example: f(x)=9x^2,
f'(x) = 18x = 0 --> x = 0
Vertex (0, 0)
OR a=9, b=0, c=0 -->
x = -b/(2a) = 0
y = 0