Calculus Help

Question

I need this question solved and I am stuck.  Let f(x)=2x+1−11e^x. Then the equation of the tangent line to the graph of f(x) at the point (0,−10) is given by y=mx+b. solve for m and b

Answer 1

To find the equation of the tangent line, we need to find the derivative of f(x).

 

f'(x) = 2 - 11e^x

 

 

Evaluate the derivative to get a constant value.

 

f'(0) = 2 - 11e⁰

       = 2 - 11(1)

       = 2 - 11

       = -9

 

This is the slope of the tangent line.  So far, the tangent line is

 

y = -9x + b

 

 

Plug in the values of the point (0, -10) to find the value of b.

 

-10 = -9(0) + b

 

-10 = 0 + b

 

-10 = b

 

 

The tangent line is

 

y = -9x - 10