Question

What is f(x) = 8x2 + 4x written in vertex form?

Answer 1

First of all, you need to complete the square of \(x\)

and this could be done as follows:

f(x)=8[x^2+0.5x]
     =8[x^2 + 0.5x + (0.5/2)^2 - (0.5/2)^2]
     =8[(x^2 + 0.5x +1/16) - 1/16]
     =8[(x+0.25)^2 - 1/16]


And our parabola has the vertex form of this:
 
     $f(x)=8(x+ \frac{1}{4})^2 - \frac{1}{2}$

and according to the chart I attached, the vertex is (-0.25,-0.5)


Tell me if there is any difficulty faces you!

Answer 2

The vertex form of $f(x) = 8x^2 + 4x$ is $f(x) = 8(x+\dfrac{1}{4} )^2-\dfrac{1}{2}$.

Given,

$f(x) = 8x^2 + 4x$.

We have to write $f(x)$ in vertex form.

Here,

$f(x) = 8x^2 + 4x$

How to write the equation in vertex form?

Take $8$ common from both the terms, we get

$f(x) = 8(x^2 + \dfrac{x}{2})$

Now, make the term of x as a perfect square,

$f(x) = 8(x^2 +2\times x\times \dfrac{1}{4} +\dfrac{1}{16} -\dfrac{1}{16} )$

$f(x) = 8((x+\dfrac{1}{4} )^2-\dfrac{1}{16} )$

$f(x) = 8(x+\dfrac{1}{4} )^2-\dfrac{1}{2}$

Hence, the vertex form of $f(x) = 8x^2 + 4x$ is $f(x) = 8(x+\dfrac{1}{4} )^2-\dfrac{1}{2}$.

For more details about vertex form, follow the link:

https://brainly.com/question/15165354