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

Jul 12, 2017

$y = \frac{1}{3} {\left(x - \left(- \frac{3}{8}\right)\right)}^{2} - \frac{67}{64} \leftarrow$ this is the vertex form.

#### Explanation:

The given equation :

$y = \frac{1}{3} {x}^{2} + \frac{1}{4} x - 1 \text{ [1]}$

Is in the standard form:

$y = a {x}^{2} + b x + c \text{ [2]}$

where $a = \frac{1}{3} , b = \frac{1}{4} , \mathmr{and} c = - 1$

The desired vertex form is:

$y = a {\left(x - h\right)}^{2} + k \text{ [3]}$

The "a" in equation [2] is the same value as the "a" in equation [3], therefore, we make that substitution:

$y = \frac{1}{3} {\left(x - h\right)}^{2} + k \text{ [4]}$

The x coordinate of the vertex, h, can found be using the values of "a" and "b" and the formula:

$h = - \frac{b}{2 a}$

Substituting in the values for "a" and "b":

$h = - \frac{\frac{1}{4}}{2 \left(\frac{1}{3}\right)}$

$h = - \frac{3}{8}$

Substitute the value for h into equation [4]:

$y = \frac{1}{3} {\left(x - \left(- \frac{3}{8}\right)\right)}^{2} + k \text{ [5]}$

The y coordinate of the vertex, k, can be found by evaluating equation [1] at $x = h = - \frac{3}{8}$

$k = \frac{1}{3} {\left(- \frac{3}{8}\right)}^{2} + \frac{1}{4} \left(- \frac{3}{8}\right) - 1$

$k = - \frac{67}{64}$

Substitute the value for k into equation [5]:

$y = \frac{1}{3} {\left(x - \left(- \frac{3}{8}\right)\right)}^{2} - \frac{67}{64} \leftarrow$ this is the vertex form.