Question

How do you graph the parabola `f(x)=x^2+2` using vertex, intercepts and additional points?

Answer 1

See the answer process below:

Explanation:

This can be written in `ax^2+bx+c` standard form:

`x^2+0x+2`

We can check for `"x-intercepts"` with the discriminant `b^2-4ac`

`0^2-4*1*2=-8`

There are no real solutions for the `"x-intercept"`.

To find the `"y-intercept(s)"`, plug in `0` for `x`.

`y=0^2+2=2`

The `"y-intercept"` is `(0,2)`.

To find the vertex, use the formula

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

`k=c-(b^2)/(4a)`

with `(h,k)` as the vertex:

`h=(-0)/(2*1)=0`

`k=2-(0^2)/(4*1)=2`

The vertex is `(0,2)`

To find additional points plug in `1 and -1` for `x:`

`1^2+2=3=>(1,3)`

`(-1)^2+2=3=>(-1,3)`

Plot these on a graph and draw the parabola.

Here is a graph for reference:

graph{x^2+2 [-10, 10, -0.24, 9.76]}