Determine the vertex of the quadratic function f(x)=2x^2 - 4x - 6 and rewrite it in vertex form

Answer 1

Hi Grant H

Given

f(x) = 2x² - 4x - 6

This is in the form of

f(x) = ax² + bx + c

where

a = 2

b = -4

c = -6

The x also called the h coordinate of the vertex = (-b/2a) = -(-4)/(2*2) =1

Plug x coordinate back into the original equation to find the y also called the k coordinate of the vertex

f(x) = 2(1)² - 4(1) - 6 = 2 - 4 -6 = -8

Coordinates of the vertex are (1, -8)

Since h = 1 and k = -8 in vertex form we

f(x) = 2(x -1)² - 8

You can graph both the vertex form and the standard form on the same grid at desmos.com to see that they represent the same parabola.

I hope this helps.

Answer 2

2x^2 -4x -6

= 2(x^2-2x +1) - 6 -2

= 2(x-1)^2 -8 is vertex form vertex = minimum point = (1,-8)

Answer 3

See the videa.

Answer 4

Vertex form is y = a(x - h)² + k.

f(x) = 2(x² - 2x) - 6 = 2(x² - 2x + 1) - 2 - 6

So, y = 2(x - 1)² - 8. Vertex is (1, -8).