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

Jun 13, 2015

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

#### Explanation:

The equation is of the kind

$y = a {x}^{2} + b x + c$ $a = - 1 , b = 0 , c = 3$ which means it is a parabola whose axis is parallel to the y axis. Notice that $a < 0$ so it is a downward facing parabola.

Now you have to calculate the vertex:

use the formula
$2 a {x}_{v} + b = 0 \implies {x}_{v} = - \frac{b}{2} a \implies {x}_{v} = 0$
${y}_{v} = a {x}_{v}^{2} + b {x}_{v} + c = 0 + 0 + 3 = 3$

So the vertex is in $\left(0 , 3\right)$, the axis is parallel to the y axis and intersect the curve in $\left(0 , 3\right) \implies$ it's the y axis $\implies$ the parabola is symmetric on the y axis.

Then you calculate the intersection with the x axis. Trivially, they are $\left(\sqrt{3} , 0\right)$ and $\left(- \sqrt{3} , 0\right)$

Now use what you know and draw the curve, as in the graph.