Posted by **Anonymous** on Sunday, November 18, 2007 at 12:40am.

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)?

- Calculus -
**Reiny**, Sunday, November 18, 2007 at 1:30am
yes

- Calculus -
**Anonymous**, Sunday, November 18, 2007 at 12:03pm
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?

- Calculus -
**Reiny**, Sunday, November 18, 2007 at 2:55pm
check your derivative

I had y' = -13/(3x-2^2

so the slope would be -13

- Calculus -
**qwerrtjdjdjswb**, Sunday, November 18, 2012 at 8:49pm
How??????

- Calculus -
**Tim Tittles**, Sunday, November 18, 2012 at 8:54pm
I don't know. Why are you asking me?

- Calculus -
**Remi**, Monday, April 15, 2013 at 9:55pm
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

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