Equation of normal of curve f(x)=x³-4x²+5x+2, at x=1

[naki12]

So I came across an interesting question in the Essential Mathematical Methods 1&2 Teacher CD-ROM:
Find the equation of the normal to the following curve f(x)=x³-4x²+5x+2 at x=1.
What's interesting is that my method of working out the normal leads me to something of a dead end.
Here is my working out:
f(x)=x³-4x²+5x+2
f(1)=1³-4*1²+5*1+2 = 4
f'(x)=3x²-8x+5
f'(1)=3*1²-8*1+5=0
here is where I am stuck, because I usually use the gradient to find the other number which times together with it to give a -1. But here, some number n*0=-1. This is not possible, thus, I cannot work out the next part to find the normal equation.

Can anyone help? Sorry for writing so much, you can just solve using the 'Subject' sentence.

[paulsterio]

This means that the tangent at that point will be a horizontal line, so the normal would be the vertical line through that point
i.e. The Normal is x = 1

[Lasercookie]

Assuming that there hasn't been any mistakes made in the working out, if the gradient is 0, that means it is a stationary point. So the gradient of the normal is -1/0, or undefined.

If the gradient is undefined, that means that it is a vertical line.

edit: my working out was completely wrong, redacted it now. y = 4 is a horizontal line - I forgot to use common sense.

[naki12]

This means that the tangent at that point will be a horizontal line, so the normal would be the vertical line through that point
i.e. The Normal is x = 1

Thanks for explanation. It is strange for the question to require an explanation instead of working out to solve; but I guess such things also exist.

[paulsterio]

There are explain questions in maths - there's one in the VCAA 2009 Exam 1 - if I remember correctly - so you need to know the theory behind it as well as how to do questions