SOLUTION: 1/16(y+4)^2 = x - 3 My teacher wants me to write this equation in standard form but I don't even know how to get started. What do I do first?

Question 610061: 1/16(y+4)^2 = x - 3
My teacher wants me to write this equation in standard form but I don't even know how to get started. What do I do first?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1/16(y+4)^2 = x - 3
My teacher wants me to write this equation in standard form but I don't even know how to get started. What do I do first?
**
First, you must recognize this is an equation of a parabola. Other conics are circles, ellipses and hyperbolas. You should be able to recognize the standard forms for each of these conics.
..
Parabolas have two basic standard forms which apply to given problem:
x=A(y-k)^2+h, (h,k)=(x,y) coordinates of the vertex
or
(y-k)^2=4p(x-h), again, (h,k)=(x,y) coordinates of the vertex
..
In this case, the second form applies, so we should rewrite given equation to look like this form:
1/16(y+4)^2 = x - 3
(y+4)^2=16(x-3) (Standard form of equation for parabola that opens rightwards)
vertex: (3,-4)
4p=16

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