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I'm trying to sketch a parabola before even finding the range of it. For some reason I've never known how to do this, whether I was away or not, I'm not sure.

So I have, $y=-x^2-2x$. I'd usually find $x$/$y$-intercepts and then the $x$-intercepts are always $2$ values in which both parts of the parabola pass through. So, I substituted $y = 0$ in to the original equation to find the $x$ values, and I'm not sure how to solve $-x^2-2x$.

Any help would be much appreciated and if possible a small walk-through for sketching parabolas based on the equation (note: I only struggle with equations that have $2-3$ terms, I'm fine with $1$).

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Not best approach, but to solve $-x^2-2x=0$, equivalently solve $x^2+2x=0$, that is, $x(x+2)=0$. This happens if $x=0$ or $x=-2$. –  André Nicolas Jul 30 '12 at 8:35
    
@AndréNicolas Alright. I get that much. Once I find that, I can find the turning point which is -1. Then we sub -1 into original equation, so, $\text{y = -(-1)^2-2(-1)}$. Then what? –  Shazer2 Jul 30 '12 at 8:41
    
For this example, what's left is to determine what type of turning point you have (minimum/maximum/point of inflection). You could also see what happens to $y$ when $x$ is very big and negative and very big and positive but that should really be clear once you've identified your turning point. –  Peter Phipps Jul 30 '12 at 10:06
    
So, you know it goes through $(-2,0)$, $(-1,1)$, and $(0,0)$. Isn't that enough information to make a sketch? –  Gerry Myerson Jul 30 '12 at 13:03

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