SOLUTION: I am a little confused in inequalities. Okay let me tell u why i am confused here.Given an equation x^2+2x+1>0. When x^2+2x+1>0 i know x^2+2x+1 is the graph. So, what is 0 here? W

Algebra ->  Graphs -> SOLUTION: I am a little confused in inequalities. Okay let me tell u why i am confused here.Given an equation x^2+2x+1>0. When x^2+2x+1>0 i know x^2+2x+1 is the graph. So, what is 0 here? W      Log On


   

Question 967735: I am a little confused in inequalities. Okay let me tell u why i am confused here.Given an equation x^2+2x+1>0. When x^2+2x+1>0 i know x^2+2x+1 is the graph.
So, what is 0 here? Whats the meaning? Does it mean y=0?
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Depends on if this is from a function or not.

Roots will be critical values for x.
x%5E2%2B2x%2B1=%28x%2B1%29%5E2
;
%28x%2B1%29%5E2%3E0 has only one intersection point with the x-axis, if your point of view is y=0=%28x%2B1%29%5E2=x%5E2%2B2x%2B1. This is a parabola, concave upward, and ABOVE the x-axis.

y=0 ONLY FOR ONE POINT and none others. What you can do is look at the intervals on x which the critical values of x=-1 create. The function is 0 at x=-1. You now examine what happens for x%3C-1; you know what happens at x=-1; and check what happens for -1%3Cx. Check the truth of the given inequality in the two intervals.


graph%28300%2C300%2C-8%2C8%2C-8%2C8%2Cx%5E2%2B2x%2B1%29