Question

How do you graph the parabola `y=-2.5(x-30)^2+40` using vertex, intercepts and additional points?

Answer 1

Axis of symmetry: `h=x=30`

Vertex: `(30,40)`

y-intercept: `(0,2210)`

x-intercepts: `(26,0),` `(34,0)`

Refer to the explanation for the process.

Explanation:

Graph the parabola:

The vertex form of a parabola is: `a(x-h)^2+k`, where the vertex is `(h,k)`.

Vertex

`y=-2.5(x-30)^2+40`,

where `h=30` and `k=40`.

The vertex is `(30,40)`, and the axis of symmetry is `x=h=30`. Because `a=-2.5`, the vertex is the maximum and the parabola opens downward.

Y-Intercept

Set `x` equal to `0`.

`y=-2.5(0-30)^2+40`

Simplify.

`y=-2.5(900)+40`

Expand.

`y=-2250+40`

Simplify.

`y=-2210`

y-intercept: `(0,-2210)`

X-Intercepts

Substitute `0` for `y` and solve for `x`.

`0=-2.5(x-30)^2+40`

Switch sides.

`2.5(x-30)^2-40=0`

Add `40` to both sides.

`2.5(x-30)^2=40`

Divide both sides by `2.5`.

`(x-30)^2=40/2.5`

Simplify.

`(x-30)^2=16`

Take the square root of both sides.

`x-30=+-4`

Add `30` to both sides.

`x=+-4+30`

`x=4+30,-4+30`

Simplify.

`x=34,` `26`

x-intercepts: `(34,0)` and `(26,0)`.

Summary

Axis of symmetry: `h=x=30`

Vertex: `(30,40)`

y-intercept: `(0,2210)`

x-intercepts: `(26,0),` `(34,0)`

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

graph{y=-2.5(x-30)^2+40 [22.84, 42.84, 31.92, 41.92]}