SOLUTION: We are doing quadratic functions to vertex form. How do I find the first number? Aka, I have the vertex, but I don't know how to find a. A beiing: Y=a(x-2/3)^2+41/3 Original p

Question 952951: We are doing quadratic functions to vertex form.
How do I find the first number? Aka, I have the vertex, but I don't know how to find a.
A beiing:
Y=a(x-2/3)^2+41/3
Original problem:
Y=3x^2-4x+15
I've tried plugging in the y-intercept of (0,15), but I get a crazy big fraction, and the website tells me its wrong!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Original problem:
Y=3x^2-4x+15
----
3(x^2-(4/3)x + ?) = y-15 + 3*?
3(x^2 - (4/3)x + (2/3)^2) = y - 15 + 3(4/9)
3(x-(2/3))^2 = y - 45/3 + 4/3
3(x-(2/3))^2 = y - (41/3)
-----
Vertex:: (2/3 , 41/3)
-------------
Cheers,
Stan H.
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