# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y= x^2 + 6x + 5?

Sep 30, 2015

The axis of symmetry is $x = - 3$.
The vertex is $\left(- 3 , - 4\right)$

#### Explanation:

$y = {x}^{2} + 6 x + 5$ is a quadratic equation which has the form $y = a {x}^{2} + b x + c$, in which $a = 1 , b = 6 , \mathmr{and} c = 5$.

Axis of Symmetry

The axis of symmetry is determined by the formula $x = \frac{- b}{2 a}$.

$x = \frac{- 6}{2 \cdot 1} = - \frac{6}{2} = - 3$

The axis of symmetry is $x = - 3$

Vertex

The vertex is the point $\left(x , y\right)$ that is the maximum or minimum of a parabola. Since $a$ is positive, the parabola for this graph will open upward and the vertex will be the minimum point.

The $x$ value of the vertex is the same as the axis of symmetry. $x = - 3$.

To find the $y$ value of the vertex, substitute $- 3$ for $x$ in the equation $y = {x}^{2} + 6 x + 5$.

$y = {\left(- 3\right)}^{2} + 6 \left(- 3\right) + 5 =$

$y = 9 - 18 + 5 = - 4$

The vertex is $\left(- 3 , - 4\right)$.

Determine several points on both sides of the axis of symmetry by substituting values for $x$ into the equation and solving for $y$.

$y = {x}^{2} + 6 x + 5$

$x = - 6 ,$ $y = 5$
$x = - 5 ,$ $y = 0$
$x = - 4 ,$ $y = - 3$
$x = - 3 ,$ $y = - 4$ (vertex)
$x = - 2 ,$ $y = - 3$
$x = - 1 ,$ $y = 0$
$x = 0 ,$ $y = 5$

Plot the points and sketch an upward opening parabola with the curve at the vertex. Do not connect the dots.

graph{y=x^2+6x+5 [-16.02, 16, -8.01, 8.01]}