Question

How do you graph `y=-(x+3)^2 + 5`?

Answer 1

Vertex: `(-3,5)`

Y-intercept: `(0,-4)`

X-intercepts: `(-3+sqrt5,0),` `(-3-sqrt5,0)`

Approximate x-intercepts: `x~~(-0.7639,0),` `"and"` `~~(-5.236,0)`

Explanation:

Graph:

`y=-(x+3)^2+5` is a quadratic equation in vertex form:

`y=a(x-h)^2+k`,

where:

`a=-1`, `h=-3`, and `k=5`

The vertex, `(h,k)`, is `(-3,5)`

To find the y-intercept, substitute `0` for `x` and solve for `y`.

`y=-1(0+3)^2+5`

`y=-9+5`

`y=-4`

Y-intercept: `(0,-4)`

To find the x-intercepts, substitute `0` for `y` and solve for `x`.

`-(x+3)^2+5=0`

`-(x+3)^2=-5`

Divide both sides by `-1`.

`(x+3)^2=(-5)/(-1)`

Simplify.

`(x+3)^2=5`

Take the square root of both sides.

`x+3=+-sqrt5`

Subtract `3` from both sides.

`x=-3+-sqrt5`

X-intercepts: `(-3+sqrt5,0),` `(-3-sqrt5,0)`

`x~~(-0.7639,0),` `"and"` `~~(-5.236,0)`

Summary:

Vertex: `(-3,5)`

Y-intercept: `(0,-4)`

X-intercepts: `(-3+sqrt5,0),` `(-3-sqrt5,0)`

Approximate x-intercepts: `x~~(-0.7639,0),` `"and"` `~~(-5.236,0)`

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

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