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

##### 1 Answer

#### Explanation:

Take the derivative of

#f^'(x) = 3(x^2)-5#

Plug

#f^(')(3) = 3(3^2)-5 = 22#

That is the slope tangent to the curve but to find the slope of the tangent line, you need to take the reciprocal of that number and then switch signs, so the slope of the normal line is

Now plug

#f(3) = 3^3-5(3)+2 = 14#

Now that we have the point

#(y-y_1) = dy/dx(x-x_1)#

Plug in

#y = (-x/22) + (45/22)#