For f(x)=x3 what is the equation of the tangent line at x=-2?

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mikedejoyacqs · Accepted Answer

Explanation:We require the slope of the tangent and a point on itslope of tangent = f&#039;(x) at x=-2f′(x)=3x2f′(−2)=3(−2)2=12f(−2)=(−2)3=−8⇒(−2,−8)← pointy+8=12(x+2)y=12x+16← equation of tangent

Reagan Blair · Accepted Answer

Explanation:Setting x=-2 in the given function we get y-coordinate of point as followsy=f(-2)=(−2)3=-8The coordinates of point are (-2, -8)Now, differentiating given function w.r.t. x, we get slope of tangent dydx as followsdydx=f′(x)=ddx(x3)=3x2hence the slope m of tangent at x=-2, is given asm=f&#039;(-2)=3(−2)2=12Now, the equation of tangent at the point (x1,y1)≡(−2,−8) &amp;amp; slope m=12is given by following formulay−y1=m(x−x1)y-(-8)=12(x-(-2))y+8=12x=2412x-y+16=0

For f(x)=x^{3} what is the equation of the tangent line

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