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Mz_P · Nov 11, 2009, 12:08 PM

hello guys I need help with graphing parabolas its hard

y=-X2+1 I need help with graphing and how to do the equation like how to solve it I don't get it at HELP!!

Unknown008 · Nov 12, 2009, 07:42 AM

1. Look at the coefficient of x^2 (x squared). If it is positive it is a parabola with a minimum, and if it is negative, it is a parabola with a maximum. The value of the coefficient determines the 'way' the slope increases.

2. Try to put your equation in the form $y = a(x-b)^2 +c$

Where (b,c) is the coordinates of your optimum.

3. Find any x or y intercepts by putting y and x to zero respectively.

In your example, you have $y = -x^2+1$

We can say that this is the equivalent of : $y=-(x-0)^2 + 1$

Looking up, you'll see that (0, 1) are the coordinates of the optimum.
Then, setting y to zero, we get (-1, 0) and (1, 0) as the x intercepts.

I hope it helped! :)

mathwiz3502 · Nov 28, 2009, 06:29 AM

for a better realization, it is a regular parabola, flipped over the x-axis, and up one unit