# Thread: Finding equation of a tangent line.

1. ## 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?

2. You found the slope of the tangent line to be 1/2 = 0.5. Let $\displaystyle y = g(t)$. Then the tangent line can be expressed as $\displaystyle y - y_1 = 0.5(t - t_1)$. But $\displaystyle t_1 = -5$ so the line is $\displaystyle y - y_1 = 0.5(t + 5)$. You have to calculate the value of the function $\displaystyle y_1 = g(t_1) = g(-5)$.

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

4. 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.