Question

How do you graph the parabola `y = x^2 -12` using vertex, intercepts and additional points?

Answer 1

Refer to the explanation.

Explanation:

Given:

`y=x^2-12`

This is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=1`, `b=0`, `c=-12`

Vertex: maximum or minimum point of the parabola.

The x-coordinate of the vertex is found using the formula for the axis of symmetry:

`x=(-b)/(2a)`

`x=0/2`

`x=0`

To find the y-coordinate of the vertex, substitute `0` for `x` and solve for `y`.

`y=0^2-12`

`y=-12`

The vertex is `(0,-12)`

Y-intercept: value of `y` when `x=0` is also `(0,12)`.

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

Substitute `0` for `y`.

`0=x^2-12`

Add `12` to both sides.

`12=x^2`

Switch sides.

`sqrt(x^2)=+-sqrt12`

Apply rule `sqrt(a^2)=a`

`x=+-sqrt12`

Prime factorize `12`.

`x=+-sqrt(2xx2xx3)`

Rewrite in exponent form.

`x=+-sqrt(2^2xx3)`

Apply rule: `sqrt(a^2)=a`

`x=+-2sqrt3`

The x-intercepts are `(2sqrt3,0)` and `(-2sqrt3,0)`.

Approximate x-intercepts: `(3.464,0)` and `(-3.464,0)`

Additional Points

Choose values for the x-coordinates, substitute for `x` in the equation, and solve for `y`.

Additional point 1

`x=2`

`y=2^2-12`

`y=4-12`

#y=-8

Point: `(2,-8)`

Additional point 2

`x=-2`

`y=(-2)^2-12`

`y=4-12`

`y=-8`

Point: `(-2,-8)`

Additional point 3

`x=4`

`y=4^2-12`

`y=16-12`

`y=4`

Point: `(4,4)`

Additional point 4

`x=-4`

`y=(-4)^2-12`

`y=16-12`

`y=4`

Point: `(-4,4)`

Summary

Vertex: `(0,-12)`

Y-intercept: `(0,-12)`

X-intercepts: `(2sqrt3,0)` and `(-2sqrt3,0)`.

Approximate x-intercepts: `(3.464,0)` and `(-3.464,0)`

Additional points: `(2,-8)`, `(-2,-8)`, `(4,4)`, `(-4,4)`

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

graph{y=x^2-12 [-14.42, 14.05, -17.48, -3.25]}