Find the equation of the tangent to the curve

Quote:

Originally Posted by johnreal

Find the equation of the tangent to the curve y = ln(3x − 2) at the point where x = 1. I differentiated this and got 3/3x-2, whats my next step?

to find the equation of a line tangent to a curve defined by $f(x)$ at $x = a$ ...

1. you need the point of tangency on the curve, $(a, f(a))$

2. you need the value of the slope of the curve at the point of tangency , $m = f'(a)$

3. now that you have a point $(a,f(a))$ and a slope, $m = f'(a)$ , write the equation in point-slope form ...

$y - f(a) = f'(a)(x - a)$