Let f(x)=4-x^2. The equation of the tangent line to the curve at (1,3) is 2x+y=5.

Question

a. What is the slope of the tangent line to the curve at (1,3)?

b. What is the derivative of the function when x=1?

Answer 1

2x+y=3

y =-2x + 3, slope = -2 = the coefficient of x when the equation is solved for y

derivative = y=4-x^2 is y' =-2x

at the point (1,3) the derivative =-2(1) = -2, the same as the slope of the tangent line