# How do you write the quadratic function y=2x^2+4x-5 in vertex form?

Jul 24, 2017

See a solution process below:

#### Explanation:

To convert a quadratic from $y = a {x}^{2} + b x + c$ form to vertex form, $y = a {\left(x - \textcolor{red}{h}\right)}^{2} + \textcolor{b l u e}{k}$, you use the process of completing the square.

First, we must isolate the $x$ terms:

$y + \textcolor{red}{5} = 2 {x}^{2} + 4 x - 5 + \textcolor{red}{5}$

$y + 5 = 2 {x}^{2} + 4 x$

We need a leading coefficient of $1$ for completing the square, so factor out the current leading coefficient of 2.

$y + 5 = 2 \left({x}^{2} + 2 x\right)$

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by $2$ on the left side of the equation. This is the coefficient we factored out in the previous step.

y + 5 + (2 * ?) = 2(x^2 + 2x + ?)

$y + 5 + \left(2 \cdot 1\right) = 2 \left({x}^{2} + 2 x + 1\right)$

$y + 5 + 2 = 2 \left({x}^{2} + 2 x + 1\right)$

$y + 7 = 2 \left({x}^{2} + 2 x + 1\right)$

Then, we need to create the square on the right hand side of the equation:

$y + 7 = 2 {\left(x + 1\right)}^{2}$

Now, isolate the $y$ term:

$y + 7 - \textcolor{red}{7} = 2 {\left(x + 1\right)}^{2} - \textcolor{red}{7}$

$y + 0 = 2 {\left(x + 1\right)}^{2} - 7$

$y = 2 {\left(x + \textcolor{red}{1}\right)}^{2} - \textcolor{b l u e}{7}$

Or, in precise form:

$y = 2 {\left(x - \textcolor{red}{\left(- 1\right)}\right)}^{2} + \textcolor{b l u e}{\left(- 7\right)}$

The vertex is: $\left(- 1 , - 7\right)$

Jul 25, 2017

$y = 2 {\left(x + 1\right)}^{2} - 7$

#### Explanation:

$y = 2 {x}^{2} + 4 x - 5$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{4}{4} = - 1$
y-coordinate of vertex:
$y \left(- 1\right) = 2 {\left(- 1\right)}^{2} + 4 \left(- 1\right) - 5 = 2 - 4 - 5 = - 7$
Vertex (-1, -7)
Vertex form of y:
$y = 2 {\left(x + 1\right)}^{2} - 7$