Question

Write f(x)=8x^2-4x+11 in vertex form

Answer 1

Answer:

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$

or

$f(x)=8(x-0.25)^{2}+10.5$

Step-by-step explanation:

we have

$f(x)=8x^{2}-4x+11$

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex is a minimum

Convert to vertex form

Factor the leading coefficient

$f(x)=8(x^{2}-\frac{1}{2}x)+11$

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

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+11-\frac{1}{2}$

$f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+\frac{21}{2}$

Rewrite as perfect squares

$f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}$ ----> equation in vertex form

or

$f(x)=8(x-0.25)^{2}+10.5$

The vertex is the point  (0.25,10.5)