Question

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

Answer 1

`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]}