Question 184593: I'm can't figure out how to create a quadratic equation out of the vertex and a point that it must pass through. All i've managed so far is some trial and error and come up with this.
y=2(x+2)²+4
From the vertex of (-2,4) and passing through (-1,8).
I have been fiddling around with it and have two ideas how to proceed.
I could change (x+2) to something else to move it along the x axis or change the first part of it to change the fatness or whichever you want to call it of the parabola. Regardless i figure there is a better way to do it other then trial and error, if it has been answered before i'm sorry I didn't see it. I'm self taught and this is the first time I really have had any problem.
-Thanks
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the general vertex form of a quadratic is y = a(x-h)^2 + k
(h,k) is the vertex, and "a" controls what you call the "fatness"
your question is basically solving for "a" __ algebra, not trial and error
y = a(x+2)^2 + 4
substituting (-1,8) ___ 8 = a(-1+2)^2 + 4
subtracting 4 ___ 4 = a
so the equation is ___ y = 4(x+2)^2 + 4
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