# How do you graph y=-x^2+4?

Apr 12, 2017

Point plotting or finding the vertex and axis of symmetry

#### Explanation:

Point plotting:
Create a table of $x$ and $y$ values.

Since $x$ is the independent variable in the equation ($y$ depends on the $x$ variables selected), you can select any $x$ value and find the corresponding $y$ value using the equation $y = - {x}^{2} + 4$

$\text{x|"-3"|"-2"|"-1"| "0"| "1"| "2"| "3"|}$
$\text{y|"-5"| "0"| "3"| "4"| "3"| "0"|"-5"|}$

Plot these points on a coordinate plane and connect the points with an arc.

Finding the vertex and axis of symmetry:

A parabola can be graphed easily when it is in the standard form/vertex form $y = a {\left(x - h\right)}^{2} + k$ where the vertex $= \left(h , k\right)$ and the axis of symmetry is $x = h$

First put the equation in general form $A {x}^{2} + B x + C = 0$

A negative $A$ value means the parabola opens downward, a positive $A$ value means the parabola opens upward.

$h = - \frac{B}{2 A}$

For $y = - {x}^{2} + 4 , \text{ " A = -1, B = 0, C = 4; " } h = \frac{0}{-} 2 = 0$

So the axis of symmetry is $x = 0$

$k = f \left(h\right) = f \left(0\right) = - {\left(0\right)}^{2} + 4 = 4$

So the vertex is $\left(0 , 4\right)$

graph{-x^2+4 [-12.66, 12.66, -6.33, 6.33]}