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

By signing up, you agree to our Terms of Service and Privacy Policy

To find the slope of the tangent line to the parabola y = 7x - x^2 at the point (1, 6), you differentiate the equation of the parabola with respect to x to find the derivative. Then, you evaluate the derivative at x = 1 to find the slope of the tangent line at that point.

The derivative of y = 7x - x^2 is dy/dx = 7 - 2x.

When x = 1, dy/dx = 7 - 2(1) = 5.

So, the slope of the tangent line to the parabola y = 7x - x^2 at the point (1, 6) is 5.

By signing up, you agree to our Terms of Service and Privacy Policy

- What is the instantaneous velocity of an object with position at time t equal to # f(t)= ((t-2)^3,sqrt(5t-3)) # at # t=2 #?
- How do you find the slope of the tangent line to the graph of the given function # y=sinx+3#; #x=pi#?
- What is a Normal Line to a Tangent?
- Use the definition of the derivative to find f’ when f(x) =6x^2 +4. No points for any other methods please help??
- What is the equation of the normal line of #f(x)=x^3/(3x^2 + 7x - 1 # at #x=-1 #?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7