# How do you find the axis of symmetry, vertex and x intercepts for y=x^2+6x+5?

Apr 12, 2017

Axis of symmetry: $x = - 3$
Vertex: $\left(- 3 , - 4\right)$
X-intercepts: $x = - 5$ and $x = - 1$

#### Explanation:

Axis of symmetry:-
Can be found by applying $x = - \frac{b}{2 a}$
In the equation $y = \textcolor{red}{1} {x}^{2} + \textcolor{b l u e}{6} x + \textcolor{g r e e n}{5} , a = 1 , b = 6 , c = 5$
So, $x = - \frac{6}{2 \left(1\right)} = \textcolor{p u r p \le}{- 3}$

Vertex:-
we substitute the $x$ value $\left(\textcolor{p u r p \le}{- 3}\right)$ we found in the axis of symmetry in the function, and we will get
${\left(- 3\right)}^{2} + 6 \left(- 3\right) + 5 = 9 - 18 + 5 = \textcolor{m a \ge n t a}{- 4}$,
The vertex is the point $\left(\textcolor{p u r p \le}{- 3} , \textcolor{m a \ge n t a}{- 4}\right)$

X-intercepts:-
X-intercept is when $y = 0$, which means ${x}^{2} + 6 x + 5$ should be equal to zero.
${x}^{2} + \textcolor{\mathmr{and} a n \ge}{6} x + \textcolor{g r e y}{5} = 0$
Now we can factor it by this way since the coefficient of ${x}^{2}$ is $1$, what two numbers if multiplied will give you $\textcolor{g r e y}{5}$ and if added will give you $\textcolor{\mathmr{and} a n \ge}{6}$? They are $\textcolor{b r o w n}{5}$ and $\textcolor{\mathrm{da} r k b l u e}{1}$, so,
$\left(x + \textcolor{b r o w n}{5}\right) \left(x + \textcolor{\mathrm{da} r k b l u e}{1}\right) = 0$
$x + 5 = 0$ and $x + 1 = 0$
$x = - 5$ and $x = - 1$