What is the equation of the line normal to f(x)=−15−x2 at x=-3?

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folytonr4qx · Accepted Answer

Explanation:Slope of the tangent of a curve, at a given point, is given by the value of the first derivative at that point. As normal is perpendicular to tangent, if m is the slope of tangent, slope of normal would be −1m.To find the point, let us put x=-3 in f(x)=−15−x2 andf(−3)=−15−(−3)2=−1−4=14.Hence, we are seeking normal at (−3,14)dfdx=−(−1(5−x2))×(−2x)=−2x(5−x2)2Hence slope of tangent at x=-3 is−2(−3)(5−(−3)2)2=6(−4)2=616=38and slope of normal is −138=−83and equation of line normal is(y−14)=−83(x−(−3))or 12y-3=-32x-96 or 32x+12y+93=0

ze2m1ingkdvu · Accepted Answer

f(x)=15−x2 and x0=−3.Find the value of the function at the given point: y0=f(−3)=−14.The 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)=(15−x2)′=2x(x2−5)2Hence, M(x0)=−1f′(x0)=−(x02−5)22x0Next, find the slope at the given point.m=M(−3)=83Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−(−14)=8(x−(−3))3Or, more simply: y=8x3+314

What is the equation of the line normal

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