What is the equation of the line that is normal to f(x)=−2x3+4x−2 at x=-2?

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prangijahnot · Accepted Answer

Explanation:The normal line is perpendicular to the tangent line.f′(x)=−6x2+4 so f&#039;(-2)=-20, so the slope of the normal line is 120,the opposite reciprocal.f(−2)=−2(−2)3+4(−2)−2=6The equation of the normal line is y−6=120(x+2)

aludirao09au · Accepted Answer

Explanation:At x=-2, the y co-ordinate of the point would be −2(−2)3+4(−2)−2=6the given point through which the required normal would pass is (-2,6)Slope of the curve is dydx=−6x2+4. The slope at point (-2,6) would be−6(−2)2+4=−20The slope of the normal passing through this point would thus be 120 (Recollect the formula for slopes of two perpendicular lines m1m2=−1)The point slope form of the required line would be y−6=120(x+2)

What is the equation of the line that

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