Question

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

Answer 1

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`

`"x|"-3"|"-2"|"-1"| "0"| "1"| "2"| "3"|"`
`"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(x - h)^2 +k` where the vertex `= (h, k)` and the axis of symmetry is `x = h`

First put the equation in general form `Ax^2 +Bx + C = 0`

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

`h = -B/(2A)`

For `y = -x^2 + 4," " A = -1, B = 0, C = 4; " " h = 0/-2 = 0`

So the axis of symmetry is `x = 0`

`k = f(h) = f(0) = -(0)^2 + 4 = 4`

So the vertex is `(0, 4)`

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