SOLUTION: 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 th

Question 184550: 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)�+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 checkley77(12844) (Show Source): You can put this solution on YOUR website!
y=2(x+2)�+4
y=2(x^2+4x+4)+4
y=2x^2+8x+8+4
y=2x^2+8x+12
(graph 300x200 pixels, x from -6 to 5, y from -10 to 10, 2x^2 +8x +12).
Using the x=-2 value we get:
y=2(-2+2)^2+4
y=2*0+4
y=4 ans.
(-2,4) is the vertex.
If x=-4 then:
y=2(-4+2)^2+4
y=2*-2^2+4
y=2*4+4
y=8+4
y=12
(-4,12) is another point on the parabola.
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