Asked by Anonymous

An equation of the line tangent to the graph of y=(2x+3)/(3x-2) at the point (1,5) is...do I take the derivative of y=(2x+3)/(3x-2) to find the slope and then plug that into y-5=m(x-1)?
Answered by Reiny
yes
Answered by Anonymous
I took the derivative and got -10/(3x-2)^2. So...I plug in 1 in the x. Then, I get the slope to be -10. Finally I plug it in (1,5) and m=-10 to get y=-10x+15. Is that correct?
Answered by Reiny
check your derivative
I had y' = -13/(3x-2^2
so the slope would be -13
Answered by qwerrtjdjdjswb
How??????
Answered by Tim Tittles
I don't know. Why are you asking me?
Answered by Remi
I believe the answer is y = -13x + 18. In the question I had, one of the choices was 13x + y = 18, which is what y = -13x + 18 equals. I got -13 for slope, then plugged it in to get y -5 = -13x + 13
Answered by Kaitlin
Y=-13+18
Answered by Tobey Maguire
the slope is about 6.28492178, so then plug and chug that into y-y1 = m(x-x1)
Answered by chicken
i got the same slope of -13. but according to the wks its wrong???
Answered by chicken
oops i meant *-18
Answered by chicken
o wait it is -13 hahah im going crazy
Answered by Anonymous
yes you take the derivative which gets you -13/(3x-2)^2
so that means the slope is -13 and you plug the slope and the point (1,5) into the tangent line equation y-y1=m(x-x1) so the final answer will be
13x+y=18
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