Question

Using the completing-the-square method, find the vertex of the function f(x) = –3x2 + 6x − 2 and indicate whether it is a minimum or a maximum and at what point.
A. Maximum at (1, 1)
B. Minimum at (1, 1)
C. Maximum at (–1, 2)
D. Minimum at (–1, 2)

Answer 1

we have

$f(x) =-3x^{2}+ 6x-2$

Since the leading coefficient is negative, the function has a maximum

Let

y=f(x)

$y=-3x^{2}+ 6x-2$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y+2=-3x^{2}+ 6x$

Factor the leading coefficient

$y+2=-3(x^{2}-2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y+2-3=-3(x^{2}-2x+1)$

$y-1=-3(x^{2}-2x+1)$

Rewrite as perfect squares

$y-1=-3(x-1)^{2}$

$y=-3(x-1)^{2}+1$

the vertex is the point $(1,1)$

therefore

the answer is the option

A. Maximum at (1, 1)