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January 30, 2015

January 30, 2015

Posted by **Michelle** on Monday, December 6, 2010 at 6:36pm.

the equation is y = (x-2)^2 - 1

when I graph it, the shape is supposed to be a parabola.

I'm using -2,-1,0,1, and 2 as my X values.

But I'm getting 15, 8, 3, 0, and -1 for my Y values, respectively.

- algebra -
**Jennifer**, Monday, December 6, 2010 at 7:45pmI'm not sure what you mean by solving it wrong, as you...don't appear to be solving it at all? Are you supposed to be finding the x-intercepts?

The graph of this function is a parabola, with vertex at (-h, k). You can find this vertex from the form y = a(x-h)^2 + k. So, in this instance, the vertex is at (2, -1) - so that point will be the lowest point on the parabola.

The numbers you have chosen are all to the left of or equal to the x value of the vertex of the parabola, which is why they all decrease and never increase. If you use 3 or a larger number as an x value, you'll see that the numbers begin increasing - at x = 3, we would have y = (3-2)^2 - 1 = 0 again, and so on.

Does that answer your question, hopefully?

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