## Calculus: Early Transcendentals (2nd Edition)

The equation of the tangent line at the given point is $y=\dfrac{1}{2}x+\dfrac{1}{2}$
$y=\dfrac{2x^{2}}{3x-1}$ $;$ $a=1$ First, evaluate the derivative of the given expression using the quotient rule: $y'=\dfrac{(3x-1)(2x^{2})'-2x^{2}(3x-1)'}{(3x-1)^{2}}=...$ $...=\dfrac{(3x-1)(4x)-2x^{2}(3)}{(3x-1)^{2}}=\dfrac{12x^{2}-4x-6x^{2}}{(3x-1)^{2}}=...$ $...=\dfrac{6x^{2}-4x}{(3x-1)^{2}}=\dfrac{2x(3x-2)}{(3x-1)^{2}}$ Substitute $x$ by $a=1$ in the derivative found to obtain the slope of the tangent line at the given point: $m_{\text{tan}}=\dfrac{2(1)[3(1)-2]}{[3(1)-1]^{2}}=\dfrac{2(1)}{2^{2}}=\dfrac{2}{4}=\dfrac{1}{2}$ Substitute $x$ by $a=1$ in the original expression to obtain the $y$-coordinate of the point given: $y=\dfrac{2(1)^{2}}{3(1)-1}=\dfrac{2}{2}=1$ The point is $(1,1)$ 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: $y-1=\dfrac{1}{2}(x-1)$ $y-1=\dfrac{1}{2}x-\dfrac{1}{2}$ $y=\dfrac{1}{2}x-\dfrac{1}{2}+1$ $y=\dfrac{1}{2}x+\dfrac{1}{2}$ The graph of both the function and the tangent line are shown the answer section.