## Calculus: Early Transcendentals (2nd Edition)

The equation of the tangent line at the given point is $y=-2x-1$
$y=x^{3}-4x^{2}+2x-1$ $;$ $a=2$ First, evaluate the derivative of the given expression: $y'=3x^{2}-4(2)x+2$ $y'=3x^{2}-8x+2$ Substitute $x$ by $a=2$ in the derivative found to obtain the slope of the tangent line at the given point: $m_{tan}=3(2)^{2}-8(2)+2=3(4)-16+2=...$ $...=12-16+2=-2$ Substitute $x$ by $a=2$ in the original expression to obtain the $y$-coordinate of the point given: $y=2^{3}-4(2)^{2}+2(2)-1=8-4(4)+4-1=...$ $...=8-16+4-1=-5$ The point is $(2,-5)$ The slope of the tangent line and a point through which it passes are now 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: $y-(-5)=-2(x-2)$ $y+5=-2x+4$ $y=-2x+4-5$ $y=-2x-1$ The graph of both the function and the tangent line are shown in the answer section.