Precal
posted by Sam .
What is the standard form of this parabola:
2x^2+16x+24y224=0
Please Help, parabolas are evil.

Divide both sides by 2 and complete the square for the x^2 binomial.
x^2 8x 12y +112 = 0
x^2 8x +16 12y +96 = 0
(x 4)^2 12(y8) = 0 
solution:you must be solved for the y of way implicit
2x^2+16x224=24y
2x^216x+224=24y
(1/12)x^2(2/3)x+(224/24)=y
and solve it
Respond to this Question
Similar Questions

Math  algebraish
hey how do we convert y=mx+b to standard form? 
Algebra 2
16x^2+16y^216x+24y3=0 How do I solve this? 
PRECAL
Find a Parabola with Vertex (3,1) Focus (4,1) Find the vertex of 4y^2+4y16x+13=0 Find the center of an ellipse 9x^2+4y^236x24y36=0 Find an ellipse with minor axis of 8 and vertices at (9,3),(7,3) 
precal help
Find a Parabola with Vertex (3,1) Focus (4,1) Find the vertex of 4y^2+4y16x+13=0 Find the center of an ellipse 9x^2+4y^236x24y36=0 Find an ellipse with minor axis of 8 and vertices at (9,3),(7,3) 
Algebra
How can I tell what direction this parabola open up 2x^2 +16x+24y224=0 We just started this in class today do I need to change this into another format 
Algebra: Parabolas
I have a math problem that requires me to find the focal length of a parabola using the information given on a graph. It is a standard horizontal parabola, the formula being y^2 = ax. The focus point is formula a = 4p. I know that … 
Parabolas in standard form!
Hello, I am having the worst time trying to solve these parabolas and putting them into xh = a(yk)^2 form. :( There are two problems that i keep doing something wrong. could someone solve them so I can have a set up for the rest … 
Parabola
3y^2+24y+72x96=0 Can anyone put this in standard form: (y  k)2 = 4p(x  h) All help is greatly appreciated! :) 
Parabola
How do I put this parabola in standard form? 
Algebra
Can you please help me with the below question: The vertex of a parabola is located at (12, 1). The parabola also passes through the point (10, 5). Write the equation of this parabola in both vertex form and standard form. Vertex …