Question

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

Answer 1

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).