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

The derivative is $y ' = 7 - 2 x$ so the slope is $y ' \left(1\right) = 7 - 2 = 5$.
If you want the equation of the tangent line, just use the fact that $y \left(1\right) = 6$ to write it as $y = 5 \left(x - 1\right) + 6 = 5 x + 1$.
In general, the derivative of $y = a {x}^{2} + b x + c$ is $y ' = 2 a x + b$.