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

Feb 16, 2018

$y = 2 {\left(x + \frac{1}{4}\right)}^{2} - 1 \frac{1}{8}$ with the vertex as $\left(- \frac{1}{4} , - 1 \frac{1}{8}\right)$

#### Explanation:

Vertex form is given by

$y = a {\left(x - h\right)}^{2} + k$ with $\left(h , k\right)$ being the vertex.

To get to vertex form, complete the square.

$y = 2 \left({x}^{2} + \frac{1}{2} x \textcolor{red}{+ \left(\frac{1}{4} ^ 2\right) - \left(\frac{1}{4} ^ 2\right)}\right) - 1$

$y = 2 {\left(x + \frac{1}{4}\right)}^{2} - \frac{1}{2} - 1$

$y = 2 {\left(x + \frac{1}{4}\right)}^{2} - 1 \frac{1}{8}$ and the vertex is $\left(- \frac{1}{4} , - 1 \frac{1}{8}\right)$

Note: If you want to just find the vertex using $a {x}^{2} + b x + c$:

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

$k = c - {b}^{2} / \left(4 a\right)$

with $\left(h , k\right)$ being the vertex:

$h = \frac{- 1}{2 \cdot 2}$

$h = - \frac{1}{4}$

$k = - 1 + \frac{{1}^{2}}{4 \cdot 2 \cdot - 1}$

$k = - 1 \frac{1}{8}$

$\left(h , k\right)$ is $\left(- \frac{1}{4} , - 1 \frac{1}{8}\right)$

If you were to find vertex form using the method previously described, then you would need a point on the graph of $2 {x}^{2} + x - 1$.

Let's say that they told you $\left(- 1 , 0\right)$ was a point on the graph.

Using the vertex as found out with the formulas above, plug into vertex form with what you have:

$y = a {\left(x + \frac{1}{4}\right)}^{2} - 1 \frac{1}{8}$

To find out $a$, plug the point $\left(- 1 , 0\right)$ into what you have:

$0 = a {\left(- 1 + \frac{1}{4}\right)}^{2} - 1 \frac{1}{8}$

$0 = \frac{9}{16} a - 1 \frac{1}{8}$

$1 \frac{1}{8} = \frac{9}{16} a$

$\frac{18}{16} = \frac{9}{16} a$

$a = 2$ giving you

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

Here is a graph for reference: graph{2x^2+x-1 [-10, 10, -2, 5]}

Hope this helps!