What is the equation of the line that is normal to f(x)=−2x2−2ex at x=-2?

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rhyclelal80j6 · Accepted Answer

f(x)=−2x2−2exf′(x)=−4x−2exf′(−2)=8−2e2← This is the gradient of the tangent.mn=−1mtan=−18−2e2=−e28e2−2f(−2)=−8−2e2=−8e2−2e2y−y1=m(x−x1)y+8e2−2e2=−e28e2−2(x+2)y+8e2−2e2=−xe28e2−2−2e28e2−2y=xe21−8e2−33e4−16e2+2e2(2e−1)(2e+1)

muthe2ulj · Accepted Answer

Given: f(x)=−2x2−2ex and x0=−2y0=f(−2)=−8−2e2The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(−2x2−2ex)′=−2(2x+ex)′=−2(2x+ex)Hence, M(x0)=−1f′(x0)=12(2x0+ex0)Next, find the slope at the given point.m=M(−2)=12(−4+e−2)Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−(−8−2e2)=12(−4+e−2)(x−(−2))Or, more simply: y=x2(−4+e−2)+2−33e4−e2+4e4

What is the equation of the line that

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