Question

How do you find the slope of the tangent line to the parabola `y=7x-x^2` at the point (1,6)?

Answer 1

The derivative is `y'=7-2x` so the slope is `y'(1)=7-2=5`.

If you want the equation of the tangent line, just use the fact that `y(1)=6` to write it as `y=5(x-1)+6=5x+1`.

In general, the derivative of `y=ax^2+bx+c` is `y'=2ax+b`.