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

1 Answer
Sep 30, 2016

Answer:

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]}