How do you find the equation of a line tangent to the function y=(x−2)(x2+1) at x=-1?

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losse9x1 · Accepted Answer

Explanation:Tangent Line to a Curvesimplifyy=x3−2x2+x−2differentiatey′=3x2−4x+1substitute for x=-1y&#039;=8the value of y&#039; is the slope of the tangent line at x=-1and to get the point which lies on the tangent we substitute with x=-1 in the functiony=-6so the point on the line is (-1, -6) and its slope is 8substitute in the following formula to get the equationy−y1=m(x−x1)y+6=8(x+1)y=8x+2

glavninamvyw · Accepted Answer

At x=-1:y=(−1−2)((−1)2+1)y=−3×2y=-6The tangent passes through (-1, -6)The gradient:y=(x−2)(x2+1)y=x3−2x2+x−2y′=3x2−4x+1At x=-1:m=3(−1)2−4(−1)+1m=3+4+1m=8Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula:y−y1=m(x−x1)y+6=8(x+1)y=8x+8-6y=8x+2

How do you find the equation of a line tangent

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