Find the equation of each line normal to the graph of y= 2x/x-1 and parallel to the line 2x-y+1=0
Please Help
2007-11-02 07:15:08 · 5 answers · asked by john 2 in Science & Mathematics ➔ Mathematics
dont know if its same as I think it is, since english mathematical terms are different from dutch ones... anyhows...
2x-y+1=0.... is that even a real formula?!?! x depends on y, so if x changes, y automatically changes, so this formula is a paradox...
and y=2x/x-1 is the same as 2-1 (x/x = 1/1=1)
r u messing with us? :O
2007-11-02 07:19:27 · answer #1 · answered by Anonymous · 0⤊ 2⤋
y = 2x/(x-1) has a vertical asymptote at x = 1 and a horizontal asymptote at y = 2. Its graph thus has two branches, one concave down from (-infinity, 1) and the other concave up from (1, infinity).
There is no line || to the line 2x-y+1 = 0 that is normal to bothe branches at the same time. There is a line that is || to the line 2x-y+1 = 0 and normal to the left branch and another line normal to the right branch.
The slope of y = 2x/(x-1) is -2/(x-1)^2 so it must be equal to -1/2 to be normal to the given line.
-2/(x-1)^2 = -1/2 = -2/4
(x-1)^2 = 4
x-1 = +/- 2
x = 1 +/- 2 = -1 and 3
y=2x+b
1 = -2 +b --> b = 3, so y = 2x+3 is one line
3= 6 + b --> b= -3 so y = 2x -3 is other line
2007-11-02 14:55:17 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
y=2x-3 and y=2x+1
2007-11-02 14:32:10 · answer #3 · answered by Dr J 5 · 0⤊ 0⤋
8
2007-11-02 14:17:36 · answer #4 · answered by Anonymous · 1⤊ 4⤋
What?Lol
try wikianswers, they're answers are straight forward, and usually always right!, i don't understand what to say to this question otherwise. sry.:] good luck.
god bless <3
2007-11-02 14:22:34 · answer #5 · answered by Shelby-Lou xx 9 3 · 0⤊ 0⤋