# How do you graph y=-3x^2+2?

Aug 25, 2015

Find the axis of symmetry and vertex. Determine points on both sides of the parabola. Sketch a curve through the points.

#### Explanation:

$y = - 3 {x}^{2} + 2$ is a quadratic equation of the form $a {x}^{2} + b x + c$, where $a = - 3 , b = 0 , \mathmr{and} c = 2$.

Because $a$ is negative, the graph is a parabola that opens downward. We need to find the axis of symmetry, which is the horizontal line that divides the parabola into two equal halves. The axis of symmetry is the value for $x$.

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

$x = 0$

The axis of symmetry is superimposed on the y-axis.

We also need to find the vertex, which is the maximum point of the parabola. The value $x = 0$ is the $x$ value for the vertex. To find the $y$ value, we substitute $0$ into the equation.

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

$y = 2$

The vertex equals $\left(0 , 2\right)$.

Now determine some points on both sides of the axis of symmetry.

$x = - 3 ,$ $y = - 25$
$x = - 2 ,$ $y = - 10$
$x = - 1 ,$ $y = - 1$
$x = 1 ,$ $y = - 1$
$x = 2 ,$ $y = - 10$
$x = 3 ,$ $y = - 25$

Plot the vertex and the other points. Draw a curve through the the points. Do not connect the dots.

graph{y=-3x^2+2 [-10, 10, -5, 5]}