Question,
The y intercept of the tangent line to a curve at any point is always
equal to the slope at that point. If the curve passes through the point
(2, 1), find the equation of the curve.
Based off this little information I have $y=y'x+b$ where $b$ would be $y'$ evaluated at the point. I'm not sure how to get this into an equation for a curve or if there is a much better approach. Thanks in advance.
Best Answer
If you fix a point $x_0$ the equation of the tangent line in $(x_0,f(x_0))$ is $$ y = f(x_0) + f'(x_0)(x-x_0) = f'(x_0) x + f(x_0) - f'(x_0) x_0 $$ hence $q=f(x_0)-f'(x_0)x_0$ is the intercept, and the equation is (call $x=x_0$, $y=f(x_0)$ $$ y' = y - y'x $$ i.e. $$ y'(1+x) = y $$ i.e. $$ \frac{y'}{y} = \frac{1}{1+x} $$ Can you solve this?