SOLUTION: y=-x^2-3x-2 Thanks, M

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Question 23330: y=-x^2-3x-2
Thanks,
M
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
What did you want to do with this, just graph it? Of course, it is a parabola, and it opens downward, because of the negative x^2 term.

There are several ways to find the vertex of this. I think the easiest way is to use a formula for the vertex that comes from the quadratic formula. Remember that? x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ Well, this formula is just the first part of this without the radical. It turns out the vertex of a parabola y+=+ax%5E2+%2Bbx+%2Bc (which opens up or down!) is always at x=-b%2F%282a%29+

In your case, y+=+-x%5E2+-3x-2, a=-1 and b=-3, so x=+-%28-3%2F%282%28-1%29%29%29+=+-3%2F2.

Now, if x=1, then
y+=+-x%5E2+-+3x-2
y=-9%2F4+%2B9%2F2-2=+-9%2F4+%2B18%2F4+-8%2F4=+1%2F4

Vertex is at (-3%2F2,1%2F4).

Check it out with the graph:
graph+%28500%2C500%2C+-6%2C6%2C-10%2C10%2C+-x%5E2+-3x+-2%29+
Does this graph look like the vertex is at (-3%2F2,1%2F4), and does the graph open down?

R^2 at SCC.