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

Jan 17, 2016

$y = \frac{1}{8} {\left(x - 3\right)}^{2} + 2$

#### Explanation:

Vertex form of a parabola:

$y = a {\left(x - h\right)}^{2} + k$

In order to make the equation resemble vertex form, factor $\frac{1}{8}$ from the first and second terms on the right hand side.

$y = \frac{1}{8} \left({x}^{2} + 6 x\right) + \frac{25}{8}$

Note: you may have trouble factoring $\frac{1}{8}$ from $\frac{3}{4} x$. The trick here is that factoring is essentially dividing, and $\frac{\frac{3}{4}}{\frac{1}{8}} = \frac{3}{4} \cdot 8 = 6$.

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 $\frac{1}{8}$, so the addition of the $9$ is actualy an addition of $\frac{9}{8}$ to the equation. To undo this, subtract $\frac{9}{8}$ from the same side of the equation.

$y = \frac{1}{8} \left({x}^{2} - 6 x + 9\right) + \frac{25}{8} - \frac{9}{8}$

Which simplifies to be

$y = \frac{1}{8} {\left(x - 3\right)}^{2} + \frac{16}{8}$

$y = \frac{1}{8} {\left(x - 3\right)}^{2} + 2$

Since the vertex of a parabola in vertex form is $\left(h , k\right)$, the vertex of this parabola should be $\left(3 , 2\right)$. We can confirm with a graph:

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