What is the equation of the tangent line of f(x)=x^2e^(x-2)-xe^x at x=4? Calculus Derivatives Tangent Line to a Curve 1 Answer Sasha P. Mar 15, 2016 y=(24e^2 - 5e^4)x+16e^4 - 80e^2 Explanation: f(x)=x^2e^(x-2)-xe^x f'(x)=2xe^(x-2)+x^2e^(x-2)-e^x-xe^x f(4) = 16e^2 - 4e^4 f'(4) = 8e^2 + 16e^2 - e^4 - 4e^4 = 24e^2 - 5e^4 tangent line eq: y=f'(x_0)x+n 16e^2 - 4e^4 = 4(24e^2 - 5e^4) + n n = 16e^2 - 4e^4 - 96e^2 + 20e^4 = 16e^4 - 80e^2 y=(24e^2 - 5e^4)x+16e^4 - 80e^2 Answer link Related questions How do you find the equation of a tangent line to a curve? How do you find the slope of the tangent line to a curve at a point? How do you find the tangent line to the curve y=x^3-9x at the point where x=1? How do you know if a line is tangent to a curve? How do you show a line is a tangent to a curve? How do you find the Tangent line to a curve by implicit differentiation? What is the slope of a line tangent to the curve 3y^2-2x^2=1? How does tangent slope relate to the slope of a line? What is the slope of a horizontal tangent line? How do you find the slope of a tangent line using secant lines? See all questions in Tangent Line to a Curve Impact of this question 1693 views around the world You can reuse this answer Creative Commons License