# Finding equation of a tangent line.

• Oct 5th 2008, 06:58 PM
excelsion
Finding equation of a tangent line.
first i had to find the derivation of a function, and i found that to be
g'(t)= 2/(t+3)^2

now i'm asked to find the equation of the tangent line when t = -5
so i figured id plug -5 into the above equation and i ended up with:

2/(-5+3)^2= 1/2
So that means i have two points an x and a y

-5,1/2

This is where im stuck how can i find a slope with this information so i can write on the tangent equation?
• Oct 5th 2008, 07:03 PM
icemanfan
You found the slope of the tangent line to be 1/2 = 0.5. Let $y = g(t)$. Then the tangent line can be expressed as $y - y_1 = 0.5(t - t_1)$. But $t_1 = -5$ so the line is $y - y_1 = 0.5(t + 5)$. You have to calculate the value of the function $y_1 = g(t_1) = g(-5)$.
• Oct 5th 2008, 07:06 PM
excelsion
g(-5) as in using the original function i had before g(t)=2/t+3 not the derivative function?
• Oct 5th 2008, 07:20 PM
icemanfan
Quote:

Originally Posted by excelsion
g(-5) as in using the original function i had before g(t)=2/t+3 not the derivative function?

Yes.