Question

How do you graph the parabola `y+4=1/8(x+3)^2` using vertex, intercepts and additional points?

Answer 1

Note: This is a very long answer.

Vertex: `(-3,-4)`

X-intercepts: `(4sqrt2-3,0),` `(-4sqrt2-3,0)`

Approximate x-intercepts: `(-8.66,0),` `(2.66,0)`

Y-intercept: `(0,-23/8)`

Approximate y-intercept: `(0,-2.88)`

Refer to the explanation for the process.

Explanation:

Given:

`y+4=1/8(x+3)^2`

Subtract `4` from both sides to change the equation into vertex form.

`y=1/8(x+3)^2-4`

The formula for the vertex form is:

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

where:

`a=1/8`, `h=-3`, and `k=-4`

Vertex: the minimum or maximum point of the parabola

The vertex is `(h,k)`.

Therefore the vertex for the given equation is `(-3,-4)`.

Since #a

X-Intercepts: values of `x` when `y=0`

Substitute `0` for `y`.

`0=1/8(x+3)^2-4`

Simplify.

`0=((x+3)^2)/8-4`

Add `4` to both sides.

`4=((x+3)^2)/8`

Multiply both sides by `8`.

`4xx8=(x+3)^2`

Simplify.

`32=(x+3)^2`

Take the square root of both sides.

`+-sqrt32=x+3`

Subtract `3` from both sides.

`+-sqrt32-3=x`

Prime factorize `32`.

`+-sqrt((2xx2)xx(2xx2)xx2)-3=x`

Simplify.

`+-4sqrt2-3`

Solutions for `x`.

`4sqrt2-3,` `-4sqrt2-3=x`

Switch sides.

`x=4sqrt2-3,` `-4sqrt2-3`

x-intercepts: `(4sqrt2-3,0),` `(-4sqrt2-3,0)`

Approximate values: `(-8.66,0),` `(2.66,0),`

Additional Point: y-intercept: value of `y` when `x=0`

Substitute `0` for `x` and solve for `y`. This will be the y-intercept.

`y+4=1/8(x+3)^2`

Subtract `4` from both sides.

`y=1/8(x+3)^2-4`

`y=1/8(0+3)^2-4`

Simplify.

`y=1/8(3^2)-4`

`y=1/8xx9-4`

`y=9/8-4`

Multiply `-4` by `8/8`.

`y=9/8-32/8`

`y=-23/8`

y-intercept: `(0,-23/8)`

Summary

Vertex: `(-3,-4)`

X-intercepts: `(4sqrt2-3,0),` `(-4sqrt2-3,0)`

Approximate values for x-intercepts: `(-8.66,0),` `(2.66,0)`

Y-intercept: `(0,-23/8)`

Approximate y-intercept: `(0,-2.88)`

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

graph{y+4=1/8(x+3)^2 [-17.64, 14.4, -6.22, 9.8]}