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

y-9=$\frac{1}{12}$ x
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 $\frac{1}{12}$. At x=0, f(x)= 9. Hence equation of normal line at this point would be y-9=$\frac{1}{12}$ x (slope-intercept from of line).