What is the equation of the line that is normal to f(x)=3(x−2)2−5x+2 at x=3?

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titemomo8gjz · Accepted Answer

Substituting x=3 in the function f(x)=3(x−2)2−5x+2, we get y-coordinate of point as followsy=f(3)=3(3−2)2−5⋅3+2=-10The slope dydx of tangent to the curve: f(x)=3(x−2)2−5x+2 is given by differentiating the function f(x) w.r.t. x as followsf′(x)=ddx(3(x−2)2−5x+2)=6(x-2)-5Now, the slope of tangent at x=3,f&#039;(3)=6(3-2)-5=1hence, the slope m of normal at (3, -10) is given asm=−1f′(3)=−11=−1Now, equation of the normal at the point (x1,y1)≡(3,−10) &amp;amp; having sloem=-1 is given by following formulay−y1=m(x−x1)y-(-10)=-1(x-3)x+y+7=0

maggionmoo · Accepted Answer

f(x)=−5x+3(x−2)2+2 and x0=3.Find the value of the function at the given point: y0=f(3)=−10The 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)=(−5x+3(x−2)2+2)′=6x−17Hence, M(x0)=−1f′(x0)=−16x0−17Next, find the slope at the given point.m=M(3)=-1Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y-(-10)=-(x-3)Or, more simply: y=-x-7

What is the equation of the line that

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