Question

What is the value of h when the function h(x)=x^2−8x+14 is converted to vertex form?

Answer 1

h=4

Explanation:

The given function is

`h(x)=x^2-8x+14`

We add and subtract the square of half the coefficient of x.

`h(x)=x^2-8x+((-8)/(2))^2-((-8)/(2))^2+14`

`h(x)=x^2-8x+(-4)^2-(-4)^2+14`

The first three terms forms a perfect square trinomial

`h(x)=(x-4)^2-16+14`

`h(x)=(x-4)^2-2`

We now compare to the vertex form;

`h(x)=a(x-h)^2+k`

We have h=4