Answer
The equation of the tangent line at the given point is $y=-\dfrac{3}{2}x+\dfrac{17}{2}$
Work Step by Step
$y=\dfrac{x+5}{x-1}$ $;$ $a=3$
First, evaluate the derivative of the given expression using the quotient rule:
$y'=\dfrac{(x-1)(x+5)'-(x+5)(x-1)'}{(x-1)^{2}}=...$
$...=\dfrac{(x-1)(1)-(x+5)(1)}{(x-1)^{2}}=\dfrac{x-1-x-5}{(x-1)^{2}}=-\dfrac{6}{(x-1)^{2}}$
Substitute $x$ by $a=3$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{\text{tan}}=-\dfrac{6}{(3-1)^{2}}=-\dfrac{6}{2^{2}}=-\dfrac{6}{4}=-\dfrac{3}{2}$
Substitute $x$ by $a=3$ in the original expression to obtain the $y$-coordinate of the point given:
$y=\dfrac{3+5}{3-1}=\dfrac{8}{2}=4$
The point is $(3,4)$
The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-4=-\dfrac{3}{2}(x-3)$
$y-4=-\dfrac{3}{2}x+\dfrac{9}{2}$
$y=-\dfrac{3}{2}x+\dfrac{9}{2}+4$
$y=-\dfrac{3}{2}x+\dfrac{17}{2}$
The graph of both the function and the tangent line are shown the answer section.
