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

Sep 30, 2016

Check the figure and the explanation

#### Explanation:

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

${x}^{2} - 2 x - 2 < 0$

We are now in the form $f \left(x\right) < 0$, and $f \left(x\right)$ is easy to graph. Having the graph of the function, it's easy to understand when $f \left(x\right)$ is greater than zero, or lesser than zero. If you remember that, given an $x$ value, $f \left(x\right)$ represents the corresponding $y$ value, we need to see wether $y = f \left(x\right)$ is positive or negative. This simply means: is $f \left(x\right)$ 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 $a {x}^{2} + b x + c$ 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 $1 \setminus \pm \setminus \sqrt{3}$, and so $f \left(x\right) < 0$ 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]}