# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y= -3x^2 + 5?

Sep 28, 2015

Axis of symmetry is $x = 0$ .
Vertex is $\left(0 , 5\right)$
The graph will be a downward opening parabola.

#### Explanation:

$y = - 3 {x}^{x} + 5$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = - 3 , b = 0 , c = 5$.

Axis of Symmetry

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

$x = \frac{0}{2 \cdot - 3} = \frac{0}{-} 6 = 0$

The axis of symmetry is $x = 0$.

Vertex

The vertex is the maximum or minimum point on the parabola. In this case, since the coefficient of ${x}^{2}$ is $- 3$, the parabola will open downward and the vertex will be the maximum point.

The $x$ value of the vertex is $0$ from the axis of symmetry.

To find the value of $y$ for the vertex, substitute $0$ for $x$ in the equation and solve for $y$.

$y = - 3 {x}^{2} \left(0\right) + 5 =$

$y = 0 + 5 = 5$

The vertex is $\left(0 , 5\right)$.

Determine several points on both sides of the axis of symmetry.

$x = - 2$, $y = - 7$
$x = - 1$, $y = 2$
$x = 0$, $y = 5$ (vertex)
$x = 1$, $y = 2$
$x = 2 ,$ $y = - 7$

Plot the points on a graph and sketch a curved parabola through the points. Do not connect the dots.

graph{y=-3x^2+5 [-16.02, 16, -8.01, 8.01]}