+118609 $$y=(x+1)^2$$ is a concave up parabola. The root is x=-1 That is the only value that will make y=0
you want to know where y>0
Well it will be greater than zero everywhere except where x=-1
You should draw the parabola to help you understand why this is so.
Formal solution
$$x \in R \qquad where \quad x \ne-1$$
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You use distribution.
I will do the ax²+bx+c first.
The letter 'a' represents any leading coefitiant. If you had a number in front of x, then a would be the number in front squared. 'c' is what they multiply out to be. 'b' is the adding of multiple sets.
Think of an example of x²+6x+9. It goes to (x+3)². You ask, "What plus what equals six but multiplys out to 9?"
You rewrite (x+1)² to (x+1)(x+1)
Distribute it in. That is when you times every number in the first set by every number in the second set.
You get x²+x+x+1. x+x=2x. This is x²+2x+1.
Therefore, a=1, b=2, and c=1.
This is hard to explain by typing it out.
+118609 $$y=(x+1)^2$$ is a concave up parabola. The root is x=-1 That is the only value that will make y=0
you want to know where y>0
Well it will be greater than zero everywhere except where x=-1
You should draw the parabola to help you understand why this is so.
Formal solution
$$x \in R \qquad where \quad x \ne-1$$
