# How do you find the equation of the tangent and normal line to the curve #y=(x+1)/(x-3) at x=2?

Aug 6, 2016

Slope of tangent is $4 x + y - 5 = 0$ and slope of normal is $x - 4 y - 14 = 0$

#### Explanation:

At $x = 2$, $y = \frac{2 + 1}{2 - 3} = \frac{3}{-} 1 = - 3$, hence, we have to find equation of the tangent and normal at curve $y = \frac{x + 1}{x - 3}$ at $\left(2 , - 3\right)$.

For finding equation, we need to find value of derivative of $y = \frac{x + 1}{x - 3}$ at $\left(2 , - 3\right)$, for which we use quotient rule.

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x - 3\right) \times 1 - \left(x + 1\right) \times 1}{x - 3} ^ 2$

= $\frac{x - 3 - x - 1}{x - 3} ^ 2 = - \frac{4}{x - 3} ^ 2$

and slope of tangent at $x = 2$ is $f ' \left(2\right) = - \frac{4}{2 - 3} ^ 2 = - 4$ and slope of normal is $- \frac{1}{-} 4 = \frac{1}{4}$.

Now using point slope form $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

Slope of tangent is $\left(y - \left(- 3\right)\right) = - 4 \left(x - 2\right)$ or $y + 3 = - 4 x + 8$ or $4 x + y - 5 = 0$

Slope of normal is $\left(y - \left(- 3\right)\right) = \frac{1}{4} \left(x - 2\right)$ or $4 y + 12 = x - 2$ or $x - 4 y - 14 = 0$

graph{(y-(x+1)/(x-3))(4x+y-5)(x-4y-14)=0 [-16.5, 23.5, -13.68, 6.32]}