Tangent line to a quadratic at a given point:
Find the equation of the line
tangent to 
at
(1,4).
The slope of the desired line will equal the value of the derivative of y at x = 1:
.
At x = 1, 
.
Hence we can use this as the slope in the slope-intercept form of the line:
Since the point of tangency is (1,4), these coordinates must satisfy the equation of the line. We can exploit this to solve for b:
We get 
.
Hence b = 3, and the equation of the line is
.