(x+4)^2 = -4(y-1)
I have written:
Vertex: (-4,1)
Focus: (-4,0)
Directrix: y=2
The parabola opens downward
The points are (-2,0) and (-6,0)
because I substitute "y=0" onto (x+4)^2 = -4(y-1)
My question is: Am I right? If so, then why am I having a hard time solving these problems? It takes me more than like fifteen minutes to graph these parabolas and I have so many strokes on my paper from erasing my errors. GRRR!! Please help me or give me an advice to solve these type of problems, so I can pass my math class. I have a B average and would like to maintain that B for my finals. :) Do not be a punk and waste your time if you're just here to troll -___-.
Find the Vertex, Focus, and Directrix of each parabola. Need a math expert or beginner to help me.?
2013-05-28 5:09 am
回答 (3)
2013-05-28 5:29 am
✔ 最佳答案
I think you are right. I, too, had to spend some time to check the answers because I had to look up the formula: (x-(-4))^2=4(-1)(y-1). h= -4, k= 1, and a= -1. Vertex: (h,k)= (-4, 1)
Focus: (h, k+a)= (-4, 1+(-1))= (-4, 0)
Directrix: y= k-a= 1-(-1)= 2; y= 2
參考: Joseph's 487 questions to Precalculus
2013-05-28 5:12 am
All your answers are correct.
2016-12-28 5:31 pm
vertex = (h, ok) =(0, 0) 4p(y-ok) = (x-h)^2 h = 0 ok = 0 at element (5, -10) -10 +0 = a ( 5 +0)^2 a = -2/5 and p = a million/(4a) = -0.625 -5/8 y = x ^2 a = -0.4 b = 0 c = 0 y = -2/5 x^2 foci = ( -b/2a , -(b^2 -4ac)/(4a)+ a million/(4a) ) foci = ( 0, -0.625) directrix is y = 0.625 the line of symmetry is x = 0
收錄日期: 2021-04-20 20:27:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130527210908AAcxguk