Question

How do you graph the parabola `y= 5/4(x+2)^2 -1` using vertex, intercepts and additional points?

Answer 1

Vertex: `(-2,-1)`

Y-intercept: `(0,5)`

X-intercepts: `(2/(sqrt5)-2,0)` `(-2/(sqrt5)-2,0)`

Explanation:

Graph:

`y=5/4(x+2)^2-1` is a quadratic equation in vertex form:

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

where:

`a=5/4`, `h=-2`, and `k=-1`

Vertex: maximum or minimum point of the parabola, `(h,k)`

`(-2,-1)`

Y-intercept: value of `y` when `x=0`

`y=5/4(0+2)^2-1`

`y=5/4(4)-1`

Cancel `4/4`.

`y=5-1`

`y=4`

y-intercept: `(0,4)`

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

Note: This is a long process.

Substitute `0` for `y`.

`0=5/4(x+2)^2-1`

Simplify.

`0=(5(x+2)^2)/4-1`

Add `1` to both sides.

`1=(5(x+2)^2)/4`

Multiply both sides by `4`.

`4xx1=(5(x+2)^2)/4xx4`

Cancel `4` on right-hand side.

`4=5(x+2)^2`

Divide both sides by `5`.

`4/5=(5(x+2)^2)/5`

Cancel `5` on the right-hand side.

`4/5=(x+2)^2`

Take the square root of both sides.

`+-sqrt(4/5)=x+2`

Simplify.

`+-(sqrt4)/(sqrt5)=x+2`

Simplify.

`+-2/(sqrt5)=x+2`

Subtract `2` from both sides.

`-2+-2/(sqrt5)=x`

Switch sides.

`x=+-2/(sqrt5)-2`

x-intercepts: `(2/(sqrt5)-2,0)` `(-2/(sqrt5)-2,0)`

Approximate x-intercepts: `(1.106,0)` `(-2.894,0)`

Summary:

Vertex: `(-2,-1)`

Y-intercept: `(0,5)`

X-intercepts: `(2/(sqrt5)-2,0)` `(-2/(sqrt5)-2,0)`

Approx. x-intercepts: `(1.106,0)` `(-2.894,0)`

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

graph{y=5/4(x+2)^2-1 [-10, 10, -5, 5]}