# Question #5eb19

Aug 17, 2017

Vertex: $\left(- 1 , 0\right)$

Axis of symmetry: $x = - 1$

$x$-intercept: $\left(- 1 , 0\right)$

#### Explanation:

The vertex is $\left(- 1 , 0\right)$. The equation is in vertex form, which looks like

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

where $h$ is the $x$-coordinate of the vertex and $k$ is the $y$-coordinate. The equation above has no $k$ value, so it's just $0$.

The axis of symmetry is $x = - 1$. The axis of symmetry is always the $x$ value of the vertex. You could also find it through the standard form of a parabola that looks like

$A {x}^{2} + B x + C = 0$

by calculating $- \frac{B}{2 A}$.

The $x$-intercept is $\left(- 1 , 0\right)$. You find this by setting $y = 0$ and solving for $x$.

The $x$-intercept is the factors of the quadratic equation getting set equal to $0$. For example, the $x$-intercepts for the quadratic

$\left(x + 4\right) \left(x - 3\right) = 0$

are $\left(- 4 , 0\right)$ and $\left(3 , 0\right)$. You get this by doing

$x + 4 = 0 \text{ }$ and $\text{ } x - 3 = 0$

and solving for $x$. For the equation above, there is only one different factor. That would only give one $x$-intercept. The graph would bounce on the point $\left(- 1 , 0\right)$ due to the even multiplicity.

graph{(x+1)^2 [-13.89, 12.96, -10.4, 48]}