The tangent line to y= f(x) at (4,-6) passes through the point (10,9) compute the following, f(4) and f'(4)

The solution to your problem is as follows:

The tangent line passes through (4,-6), so x = 4 when y = -6 

=> f(4) = -6 

The tangent line will have a constant gradient. 

gradient is m = (-6 - 9)/(4 - 10) => 5/2

Equation is y - 9 = (5/2)(x - 10) 

=> y - 9 = (5/2)(x-10) 

i.e. y = (5/2)(x-10) + 9 
      y = (5/2)x - 25 + 9 =   y = (5/2)x - 16

Now, f '(x) = dy/dx = 5/2

so, f '(4) = 5/2i.e. gradient is 5/2 whatever the value of x. 

I hope my answer has come to your help. Thank you for posting your question here in .