Answer:
The vertex for the function f(x) = 3(x – 2)2 + 4 is at (2, 4).
Explanation:
Find the vertex for f(x) = 3 (x - 2)^2 + 4
f(x) = 3 (x - 2)^2 + 4 can also be written as:
y = 3 (x - 2)^2 + 4
To find critical points, first compute f'(x):
d/(dx)(3 (x - 2)^2 + 4) = 6 (x - 2):
f'(x) = 6 (x - 2)
Solve 6 (x - 2) = 0
6x - 12 = 0
6x = 12
x = 2
iI you substitute x = 2 in 3 (x - 2)^2 + 4 then you get:
y = 3 (x - 2)^2 + 4
x = 2
y = 3 (2 - 2)^2 + 4
y = 3 (0)^2 + 4
y = 3 (0) + 4
y = 4
Answer: The vertex for the function f(x) = 3(x – 2)2 + 4 is at ( 2, 4 ).