Question

Find the slope and the equation of the line tangent to f (x )equals StartFraction 2 x minus 1 Over x plus 7 EndFraction at x​ = 2. The slope of the line tangent to​ f(x) at x​ = 2 is nothing. Answer the following in​ slope-intercept form. ​ (y = mx​ + b) The equation of the tangent line is y​ = nothing

Answer 1

`y=(5)/(27) x-(1)/(27)`

Explanation:

`f(x)=(2x-1)/(x+7)`

To find slope of f(x) at x=2, find the derivative f'(x)

apply quotient rule to find derivative

`f(x)=(2x-1)/(x+7)\\f'(x)=(2(x+7)-1(2x-1))/((x+7)^2) \\f'(x)=(15)/((x+7)^2)`

f'(x) is the slope . Now find slope at x=2. plug in 2 for x

`f'(x)=(15)/((x+7)^2)\\f'(2)=(15)/((2+7)^2)=(5)/(27)`

find out f(x) when x=2

`f(x)=(2x-1)/(x+7)\\f(2)=(2(2)-1)/(2+7)=(1)/(3)`

Now frame the equation of the line

(2,1/3) slope = 5/27

`y-y_1=m(x-x_1)\\y-(1)/(3)=(5)/(27) (x-2)\\y-(1)/(3)=(5)/(27) x-(10)/(27)\\\\y=(5)/(27) x-(1)/(27)\\\\`