Question

Find an equation of the tangent line to the curve at the given
point.

Answer 1

\[y=\ln (x^2-3x+1) at (3,0)\]

Answer 2

I know how to take the derivative and all , I just don't knw how to put it in the form \[y-y_1 = m(x-x_1)\] and where to plug in (3,0)

Answer 3

for your equation, you put the coordinates of the point where you determine the tangent line x1 = 3 and y1 = 0, so y-0 = m(x-3)

Answer 4

yea, i think Shi is right...
if y=ln(f(x)), so
m=y'=f'(x)/f(x)
put x to m

Answer 5

so since the derivative is \[(2x-3)/(x^2-3+1) \] I need to multiply this by (x-3)?

Answer 6

no, the derivative gives you the slope of a function at any point x. For the tangent line at (3,0), like RadEn said, you plug in x = 3, and that will give you the slope of the tangent line, m

Answer 7

thank you both :) i got it ...