What is the equation of the normal line of f(x)=2x3−x2−3x+9 at x=-1?

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ze2m1ingkdvu · Accepted Answer

f(x)=2(−1)3−(−1)2−3(−1)+9f(x)=-2-1+3+9f(x)=9Therefore, the tangent passes through (-1, 9).Now that we know this, we must differentiate the function.By the power rule:f′(x)=6x2−2x−3The slope of the tangent is given by evaluating f(a) inside the derivative, a beingx=a.f′(−1)=6(−1)2−2(−1)−3f&#039;(-1)=6+2-3f&#039;(-1)=5The slope of the tangent is 5. The normal is always perpendicular to the tangent, so the slope will be the negative reciprocal of that of the tangent.This means the slope of the normal line is −15. By point slope form, we can find the equation of the normal line:y−y1=m(x−x1)y−9=−15(x−(−1))y−9=−15x−15y=−15x−15+9y=−15x+445∴ The equation of the normal line is y=−15x+445.

What is the equation of the normal line

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