Finding the equation of the tangent line to the curve $(3x+1)/(7x-1)$ at any point $P$

Question

So, the question is to find the tangent line to the curve
$\ (3x+1) / (7x - 1) $ at any specific point P.

So I start by derivating the function and get: $\ -10 / (7x - 1)^2 $.

I now know that I'm supposed to use the formula: Y - Y0 = K(X - X0) to get the function. However, am I just supposed to pick a random point, probably by choosing some X-value, on the curve or is it important which point i use?

Answer 1

Note that $K=f'(x)$ at the given point where you are looking for the tangent.

Note that for the given function

$$f'(x)= \frac{-10}{(7 x -1)^2}$$

If you don't have a specific point the general form for the tangent at $P(x_0,y_0)$ is

$$(y-y_0)=\frac{-10}{(7 x_0 - 1)^2}(x-x_0)$$

Answer 2

In general the tangent lines at different points of a function are different lines, so it has little sense to speak of a tangent to a curve without specify a point of tangency, or al least a point where the line have to passes.

In general, fixed the value of $x=a$ in the domain, we find a point of the graph of the function, $P=(a,f(a))$, that in your case is $$ f(a)=\frac{3a+1}{7a-1} $$ and the slope of the tangent is the value of the derivative at this point $K=f'(a)$, in your case: $$K=\frac{-10}{(7a-1)^2}$$

So the tangent line has equation: $$ y-f(a)=K(x-a) $$