# How do you graph and label the vertex and axis of symmetry y=x^2+2x?

Sep 29, 2017

See below.

#### Explanation:

To graph we need to find significant points.

The y axis intercept occurs where $x = 0$:

So: $y = {0}^{2} + 2 \left(0\right) \implies y = 0$

$x$ axis intercepts occurs where $y = 0$

Factor ${x}^{2} + 2 x$

$\left({x}^{2} + 2 x\right) \implies x \left(x + 2\right)$

$x \left(x + 2\right) = 0 \implies x = 0 , x = - 2$

So parabola intercepts $x$ axis at the points:

$\left(0 , 0\right)$ and $\left(- 2 , 0\right)$

Since you need to label vertex and axis of symmetry you will need to arrange $y = {x}^{2} + 2 x$ into the form:

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

Where $h$ is the axis of symmetry and $k$ is the maximum/minimum value.
To do this place a bracket around the terms containing the variable.

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

Factor out the coefficient of ${x}^{2}$ if this is not 1:

Add the square of half the coefficient of $x$ inside the bracket, and subtract it outside the bracket:

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

Make $\left({x}^{2} + 2 x + {\left(1\right)}^{2}\right)$ into the square of a binomial:

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

We now have:

${\left(x + 1\right)}^{2} - {\left(1\right)}^{1} \implies {\left(x + 1\right)}^{2} - 1$

We can now see that the axis of symmetry is $- 1$

Minimum value is $- 1$

We know this is a minimum value because the coefficient of ${x}^{2}$ was positive, so the parabola will be this way up $\bigcup$

Since the coordinates of the vertex is ( axis of symmetry, max/min value).

We have:

$\left(- 1 , - 1\right)$

Graph of $y = {x}^{2} + 2 x$

graph{y=x^2+2x [-5, 5, -3, 5.625]}