"Find the point(s) on the parabola $\displaystyle f(x) = x^2 - 4x + 3$ at which the normal line passes through the point (2,0). Illustrate by graphing the parabola and the normal line(s)."

I started off by finding $\displaystyle f'(x) = 2x - 4$ which would be the slope of a tangent line at x.

Next, I assumed the negative reciprocal $\displaystyle (-1)/(2x - 4)$ is the slope of the normal line.

*** Here's the problem: the tangent line's slope is simply 2, so its reciprocal is only -1/2 and not all the mess you wrote!

Remember: slope is a number.

Tonio

Then, using the given point:

$\displaystyle y - 0 = ((-1)/(2x - 4))(x - 2)$

$\displaystyle y = (-0.5)$

$\displaystyle x^2 - 4x + 3 = (-0.5)$

$\displaystyle x^2 - 4x + 3.5$

But when I use the quadratic formula I get:

$\displaystyle (4 + sqrt(2))/2$ or $\displaystyle (4 - sqrt(2))/2$

which appears to be a problem. I must have messed up somewhere...

Any help with solving this problem is appreciated.