# What is the equation of the line normal to  f(x)=(x-1)^2/(x^2+2)  at  x=1?

Dec 18, 2017

x=1

#### Explanation:

By definition, the normal line must have a slope that is the opposite reciprocal of the slope of the tangent line at $x = 1$.
Thus, to determine the slope of the normal line, we must first calculate the slope of the tangent line at x=1, which is just the derivative of $f$ at $x = 1$ or $f ' \left(1\right)$.

Step 1: Calculate $\frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{d}{\mathrm{dx}} \left[{\left(x - 1\right)}^{2} / \left({x}^{2} + 2\right)\right]$
$= \frac{2 \left({x}^{2} + 2\right) \left(x - 1\right) - {\left(x - 1\right)}^{2} \left(2 x\right)}{{x}^{2} + 2} ^ 2$

Step 2: Find f'(1)
$f ' \left(1\right) = \frac{2 \left(3\right) \left(0\right) - {\left(0\right)}^{2} \left(2\right)}{{1}^{2} + 2} ^ 3$
$f ' \left(1\right) = 0$

Step 3: Determine the equation of the normal line
Knowing that $f ' \left(1\right) = 0$ tells us that there is the graph of $f$ has a horizontal tangent line at $x = 1$. Thus, the normal line must have a slope of $\infty$, which is undefined. Because vertical lines are the only type of line with undefined slopes, $f$ must have a vertical tangent line at $x = 1$.

The equation for the vertical tangent line is $x = 1$.