Question

How many tangent lines to the curve y =x/x + 1
pass through the point (1,2)?
At which points do these tangent lines touch the curve?

Answer 1

The derivative is equal to the slope of the tangent line.

Answer 2

so once i get the slope, how to i find out how many lines?

Answer 3

so the slope is one?

Answer 4

I'm out.

Answer 5

i just don't undetstand your explanation ....

Answer 6

the derivative of x/x-1 is 1 right??

Answer 7

\[y =\frac{x}{x + 1}\]\[y'\ne1\]

Answer 8

then what does it equal?

Answer 9

you have to either use the quotient rule, of modify this to use the product rule; personally, i perfer the product rule

Answer 10

*or modify this ...

Answer 11

so it's 1/x^2+2x+1

Answer 12

\[y =\frac{x}{x + 1}\]
\[y' =\frac{(x+1)x'-x(x+1)'}{(x + 1)^2}\]
\[y' =\frac{x+1-x}{(x+1)^2}\]
\[y' =\frac{1}{(x+1)^2}\]
i agree

Answer 13

okay so now what lol

Answer 14

hmmm, a simple approach is to define any point along the curve as (a,b) and replace the xs with "a"s. and this would define the general slope of any tangent line of the curve

Answer 15

so i already have the point 1,2 so i just sub it in?

Answer 16

\[y'=m=\frac{1}{(a+1)^2}\]
\[y-b=m(x-a)\]
using the point (1,2) we can define this in terms of (a,b)
\[2=\frac{1}{(a+1)^2}(1-a)+b\]
\[2=\frac{(1-a)}{(a+1)^2}+b\]
\[2=-\frac{(a+1)}{(a+1)^2}+b\]
\[2=-\frac{1}{a+1}+b\]

Answer 17

got a mismath in the last few steps

Answer 18

what does all that mean?

Answer 19

i got y=x-1/(x+1)^2 + 2

Answer 20

keep typing your replay, i have to sign off for an hour and a half, i have math class haha. but please, keep typing! thanks a lot for your help so far, i really appericate it.

Answer 21

so the equation we get for the general tangent lines, and im going to define a,b into it counter parts of x,y
\[2=\frac{(1-a)}{(a+1)^2}+b\]\[2=\frac{(1-x)}{(x+1)^2}+y\]
and the given equation\[y=\frac{x}{x+1}\]

when we set these 2 equations equal to each other, their solutions should give us all the points that we are looking for an answer