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

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

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Answer 1 · 2016-01-06T17:32:45

y-9=#1/12# x

Explanation:

For given f(x), f '(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 #1/12#. At x=0, f(x)= 9. Hence equation of normal line at this point would be y-9=#1/12# x (slope-intercept from of line).

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

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