How do you put y+3x^2 -5x +3 into standard form for conic sections?
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I got this assigned as one of my math problems for homework, and my textbook doesn't give any examples for this kind of conic section. I know its suppose to be a parabola but i can't get much more than that, please help. Thanks.
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Answer:
There are two or three different "standard forms" for parabolas. 1) Y= ax^2+bx+c: Y= 3x^2-5x+3. Is this what you meant? OR 2) vertex form: Rewrite: move the constant term: y-3= 3x^2-5x Then factor out the 3 Y-3= 3(x^2-5/3 x) To complete the square, find (1/2)(5/3)= 5/6; (5/6)^2= 25/36; add inside and subtract outside, but remember to multiply by 3. Y-3= 3(x^2-5/3 x + 25/36)-3*25/36 Factor: Y-3= 3(x-5/6)^2 -25/12 Y= 3(x-5/6)^2-25/12 + 3 Y= 3(x-5/6)^2+ 11/12 OR....using the same steps above. 3) y-k= (1/4p)(x-h)^2 Y-3= (3)(x-5/6)^2 OR 4) (x-h)^2= 4p(y-k) (x-5/6)^2= (1/3)(y-3) Hoping this helps!
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