# What is the vertex form of #y=1/8x^2+3/4x +25/8#?

##### 1 Answer

#### Explanation:

**Vertex form of a parabola:**

#y=a(x-h)^2+k#

In order to make the equation resemble vertex form, factor

#y=1/8(x^2+6x)+25/8#

**Note:** you may have trouble factoring

Now, complete the square in the parenthesized terms.

#y=1/8(x^2+6x+9)+28/5+?#

We know that we will have to balance the equation since a

#y=1/8(x^2-6x+9)+25/8-9/8#

Which simplifies to be

#y=1/8(x-3)^2+16/8#

#y=1/8(x-3)^2+2#

Since the vertex of a parabola in vertex form is

graph{1/8x^2+3/4x +25/8 [-16.98, 11.5, -3.98, 10.26]}