Question

Find the equation of line(s) tangent to y=(x-4)/(x-1) and parallel to y=3x-7. I know m=3
and f'(x)=-3/(x-1)^2 then it's set equal to 3 and after that I'm stumped. My book is not helping

f'(x)=(1)(x-4)-(x-1)(1)/(x-1)^2
f'(x)=x-4-x+1/(x-1)^2
f'(x)=-3/(x-1)^2
so -3/(x-1)^2=3 then????

Stumped

Answer 1

solve for x.

Answer 2

Set the derivative equal to the slope of the parallel line

Answer 3

First lets heck t see if your derivative is correct.

Answer 4

f'(x)=3/(x-1)^2

Answer 5

alrighty..

Answer 6

now proceed..equate the derivative to 3 and solve for x.

Answer 7

I'm not clear on how to solve for x - my algebra is rusty

Answer 8

\[\large \frac{3}{(x-1)^2}=3\]\[\large 3=3(x-1)^2\]divide by 3 on both sides \[\large \frac{3}{3}=\frac{3(x-1)^2}{3}\]\[\large 1=(x-1)^2\]