Question

Write the equation of the line tangent to the given point. Please show steps
y=(2x+1)/(3x-1); (1,(3/2))

Answer 1

y=6x^2+x-1
y'=12x+1
So the slope of the tangent at the point (1,3/2) is 13
So now we need the equation of the line whose slope is 13 and which passes through the point (1,3/2). You can find that, can't you?

Answer 2

umm theres a division sign in the middle...

Answer 3

Ok. Let me rephrase that: So the slope of the tangent at the point (1,1.5) is 13
So now we need the equation of the line whose slope is 13 and which passes through the point (1,1.5). You can find that, can't you?

Answer 4

youre still not getting it. just nevermind.

Answer 5

Oh. I see. It's a rational function. See what happens when you get old?

Answer 6

Let's try again.

Answer 7

lol

Answer 8

\[y'=\frac{(3x-1)(2)-(2x+1)(3)}{(3x-1)^2}\]

Answer 9

That is:
\[\frac{-5}{(3x-1)^2}\]

Answer 10

Now if we let x = 1, we see that the slope of the tangent is -5/4

Answer 11

how did you get -5? i think that is where im messing up because I Keep getting 1.

Answer 12

So:
\[y-\frac{3}{2}=-\frac{5}{4}(x-1)\]

Answer 13

\[(3x-1)(2)-(2x+1)(3)=6x-2-6x-3=0-2-3=-5\]

Answer 14

ohh ok and how did that become over 4?

Answer 15

Because the denominator of the derivative is (3x-1)^2. When we find the slope at x = 1, we have[3(1)-1]^2 which is (3-1)^2 which is 2^2 which is 4

Answer 16

oh my goodness i get it. thanks sooo much!

Answer 17

yw