Question

How do you solve and graph `2x+2>x^2`?

Answer 1

Check the figure and the explanation

Explanation:

Bring everything to one side, e.g. the right one:

`x^2-2x-2<0`

We are now in the form `f(x)<0`, and `f(x)` is easy to graph. Having the graph of the function, it's easy to understand when `f(x)` is greater than zero, or lesser than zero. If you remember that, given an `x` value, `f(x)` represents the corresponding `y` value, we need to see wether `y=f(x)` is positive or negative. This simply means: is `f(x)` above or below the `x` axis?

So, draw your parabola, and only choose the points in which the parabola is negative, i.e. below the `x` axis.

In general, every parabola `ax^2+bx+c` with `a>0` has a U form, so it is negative between its two solutions, if there are two of them.

In your case, the solutions are `1\pm\sqrt(3)`, and so `f(x)<0` will be true if `1-sqrt(3) < x <1+sqrt(3)`:
this is the function
graph{x^2-2x-2 [-3.262, 7.84, -3.24, 2.31]}
and this is the region where the function is negative
graph{(x^2 - 2*x- 2) < 0 [-4.934, 4.933, -2.465, 2.468]}