What is the equation of the line that is normal to f(x)=(x−3)2+x2−6x at x=0?

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carlosegundoacyg · Accepted Answer

Explanation:For given f(x), f &#039;(x)= 2(x-3) +2x-6. Slope of tangent to f(x), at x=0 would be = -12. Therefore slope of the normal line would be 112. At x=0, f(x)= 9. Hence equation of normal line at this point would be y−9=112x (slope-intercept from of line).Result:y−9=112x

WigwrannyErarmbmk · Accepted Answer

Given:f(x)=x2−6x+(x−3)2 and x0=0.Find the value of the function at the given point: y0=f(0)=9.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)=(x2−6x+(x−3)2)′=4(x−3)Hence, M(x0)=−1f′(x0)=−14(x0−3)Next, find the slope at the given point.m=M(0)=112Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−9=x−012Or, more simply: y=x12+9

What is the equation of the line that

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