the equation of the line tangent to the curve y=(kx+8)/(k+x)

Question

at x= -2 is y=x+4
what is the value of constant k? show work please

Answer 1

The function y = (kx + 8)/(k + x)

Differentiating on each side with respect to x .

y' = [ (k + x)(k) - (kx + 8)(1)]/(k + x)²

y' = (k² + kx - kx - 8)/(k² + x² + 2kx)

y' = (k² - 8)/(k + x)²

Substitute x = - 2 in y'.

y' = (k² - 8)/[k + (- 2)]²

y' = (k² - 8)/(k² + 4 - 4k)

This is the slope of tangent line to the curve at tangent point.

 

Tangent line is y = x + 4

Slope of the above line (m) = 1

 

1 = (k² - 8)/(k² + 4 - 4k)

k² + 4 - 4k = k² - 8

k² + 4 - 4k - k² + 8 = 0

- 4k + 12 = 0

- 4k = -12

k = 3

Solution k = 3.