Finding slope and equation of tangent line

[kjones]

I need assistance.

The question reads:

Find the slope and equation of the tangent line to the graph of:

. . .f(x) = (x - 1) / (3 + x)

...at the point (-1, -1).

Plot the graph of the function and tangent line on the same diagram.

Thanks for your help. Please explain in detail for me.

K

 

[skeeter]

\(\displaystyle f(x) = \frac{x-1}{3+x}\)

to find the tangent line, you need two things ...
1. the point of tangency ... that's easy, because it was given to you (-1, -1)
2. the slope at that point of tangency ... the slope will be the value of the derivative at
x = -1. in other words ... slope, m = f'(-1). This means you'll have to find f'(x) first.

once you have the two, use the point-slope form of a linear equation to find the equation of a tangent line ... y - y₁ = m(x - x₁).

go for it.

 

[kjones]

THANKS!

K

 

[kjones]

To find the f ' (x) ... can you help me there? I can do the point slope part.

Thanks again,
K

 

[skeeter]

quotient rule