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

1 Answer
KillerBunny
Sep 30, 2016

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]}

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