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

Answer:

$y=\frac{5}{27} x-\frac{1}{27}$

Step-by-step explanation:

$f(x)=\frac{2x-1}{x+7}$

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

apply quotient rule to find derivative

$f(x)=\frac{2x-1}{x+7}\\f'(x)=\frac{2(x+7)-1(2x-1)}{(x+7)^2} \\f'(x)=\frac{15}{(x+7)^2}$

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

$f'(x)=\frac{15}{(x+7)^2}\\f'(2)=\frac{15}{(2+7)^2}=\frac{5}{27}$

find out f(x) when x=2

$f(x)=\frac{2x-1}{x+7}\\f(2)=\frac{2(2)-1}{2+7}=\frac{1}{3}$

Now frame the equation of the line

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

$y-y_1=m(x-x_1)\\y-\frac{1}{3}=\frac{5}{27} (x-2)\\y-\frac{1}{3}=\frac{5}{27} x-\frac{10}{27}\\\\y=\frac{5}{27} x-\frac{1}{27}$