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

1 Answer
Jan 17, 2016

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

Explanation:

Vertex form of a parabola:

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


In order to make the equation resemble vertex form, factor #1/8# from the first and second terms on the right hand side.

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

Note: you may have trouble factoring #1/8# from #3/4x#. The trick here is that factoring is essentially dividing, and #(3/4)/(1/8)=3/4*8=6#.

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 #9# cannot be added within the parentheses without it being counterbalanced. However, the #9# is being multiplied by #1/8#, so the addition of the #9# is actualy an addition of #9/8# to the equation. To undo this, subtract #9/8# from the same side of the equation.

#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 #(h,k)#, the vertex of this parabola should be #(3,2)#. We can confirm with a graph:

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