does anybody know anything about graphing a parabola? y=-1/4 x^2
well since it's negative it goes down
i have to plot the vertex and 4 additional points..
since there's nothing subtracted from x, it's centered vertically (symmetrical about the Y axis)
and since it's 1/4 that means it's wider than a "standard" parabola
so the vertex would be center?
yes.
what would i need to do to find the other points?
You can make as many points as you want by picking any value for X, plugging it into the equation, and getting out a value for Y.
Here there is a program where you type in the equation and the graph pops up, its free and I use it all the time. It is called Geogebra, search it on google, you will find it.
So I'd pick "easy" values for X like 1, -1, 2, -2, etc.
so then i would just type any of those in and solve?
right
there are infinitely many points on a parabola, and so if you put *any* value in for X and solve, you'll get the corresponding Y value at that point.
Let me post an example.
This is something I made with geogebra a while ago for someone who had questions about quadratic equations previously.
Here we go:
on that map where would i know what the points are ?
Wolfram alpha is also easy to use for online graphing. Here's what our parabola looks like: http://www.wolframalpha.com/input/?i=-%281%2F4%29x%5E2
You would have to look which go on there perfectly like for example: (4, -4)
so basically its wherever 2 points meet perfectltly
Yup, you try to look for one, and tell me what point you think
kk one minute
(4,-4)
Ok what about another one?
(5,-6)
Correct :)
(2,1)
2,-1
Yup thats another one
actually -2,-1
Well actually both are
the side ways number is always first then verticle right?
yes. the first number is X - horizontal/sideways. The second number is Y -- vertical/up-and-down
when i do it for my work it looked all crazy hold on going to try and attach the file
Oki doki
it wont let me post it hmnnn
yay got it right guys thank you
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