Question

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

Answer 1

`f(x)=8(x-(1)/(4))^(2)+(21)/(2)`

or

`f(x)=8(x-0.25)^(2)+10.5`

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)-(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)-(1)/(2)x+(1)/(16))+11-(1)/(2)`

`f(x)=8(x^(2)-(1)/(2)x+(1)/(16))+(21)/(2)`

Rewrite as perfect squares

`f(x)=8(x-(1)/(4))^(2)+(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)