Completing the square to find the vertex

Question

f(x)=-3x²-2+1

I tried it and I got (-3x²-2x+1)-6 after completing the square, then I factored it and got (-3x+1)(x+1)-6 , if I did it correctly how am I supposed to get a vertex from that? Or did I factor it wrong? Is it possible to get a vertex from this function?

Answer 1

f(x) = -3x² - 2x + 1                                   factor out -3

f(x) = -3(x² + (2/3)x    ) + 1                       complete the square

f(x) = -3(x² + (2/3)x + (1/3)² ) + 1 + 1/3

f(x) = -3(x + 1/3)² + 4/3

 

Vertex: (-1/3, 4/3)

Answer 2

A function in vertex form is

 

f(x) = a(x - h)² + k

 

where:

a is the coefficient of the x² term

vertex is the coordinate (h, k)

 

 

When completing the square, you want to have a term such that you get (x ± c)² , where c is a constant.

 

f(x) = -3(x² + (2/3)x + (1/9)) + (4/3)

 

f(x) = -3(x + 1/3)(x + 1/3) + (4/3)

 

f(x) = -3(x + 1/3)² + (4/3)

 

 

When you expand the vertex form, you get back the standard form of the function.

 

 

Use the vertex form of this function to find the vertex.

Answer 3

I'm going to assume you meant

 

f(x) = -3x²-2x+1

 

Completing the square results in

 

(-2*x-3*x²)+((-2*1)/(2*(-3*1)))²*(-3*1)=(0-1)+((-2*1)/(2*(-3*1)))²*(-3*1)

 

-3x²-2x-3/9=-3/9-1

 

-3x²-2x-1/3=-4/3

 

f(x) = 4/3 -3(x+1/3)²

 

The vertex for this parabola is found at x = -1/3.