Question

Consider the function f, where f(x) = 2x^2 + 6x - 8

What is the vertex form of f(x) ?

A. f(x) = 2(x - 3)^2 - 4

B. f(x) = 2(x + 3)^2 - 4

C. f(x) = 2(x - 1.5)^2 - 1.25

D. f(x) = 2(x + 1.5)^2 - 12.5

Answer 1

I wanna say B. but if i'm wrong correct me

Answer 2

The vertex form $f(x)$ is "$2(x+1.5)^2-12.5$".

Vertex form:

$\to \bold{f(x) = 2x^2 + 6x - 8}$

solving the given function:

$\to 2x^2+6x-8\\\\\to (2x^2+6x)-8\\\\\to 2(x^2+3x)-8$

taking the half of $3$ and then square it:

$\to (\frac{3}{2})^2=\frac{9}{4}$

adding $\frac{9}{4}\ \ to\ \ x^2+3x$ but remember to double the value then:

$\to 2(x^2+3x+[\frac{9}{4}])$ so far to offset adding $2\times (\frac{9}{4})$, and subtracting the $2\times (\frac{9}{4})$ from $-8$:

$\to 2(x+[\frac{3}{2}])^2-8-[\frac{18}{4}]\\\\\to 2(x+1.5)^2-8-4.5\\\\\to 2(x+1.5)^2-12.5$

Therefore, the vertex of f(x) is "Option D".

Find out more information about the function vertex here:

brainly.com/question/24625978