# How do you find the equation of the tangent and normal line to the curve y=9x^-1 at x-3?

Oct 31, 2016

I'm not sure if you meant $x = 3$ or $x = - 3$

If $x = 3$ Then
Tangent is $y = - x + 6$ and Normal is $y = x$

If $x = - 3$ Then
Tangent is $y = - x - 6$ and Normal is $y = x$

#### Explanation:

I'm not sure if you meant $x = 3$ or $x = - 3$ so let's do both:

First we need the derivative, which will give us the slope of the tangent at any point:

$y = \frac{9}{x} = 9 {x}^{-} 1 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 9 \left(- {x}^{-} 2\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{9}{x} ^ 2$

We will use the formula $y - {y}_{1} = m \left(x - {x}_{1}\right)$ to find the tangents and normals, and use the fact that the tangents and normals are [perpendicular (so their product is -1)

When x=3
 x=3 => y=3 ; y'=-9/9=-1

Tangent:
Passes through $\left(3 , 3\right)$ and has slope $m = - 1$
$\therefore y - 3 = - 1 \left(x - 3\right)$
$\therefore y - 3 = - x + 3$
$\therefore y = - x + 6$

Normal:
Passes through $\left(3 , 3\right)$ and has slope $m = 1$
$\therefore y - 3 = 1 \left(x - 3\right)$
$\therefore y - 3 = x - 3$
$\therefore y = x$

When x=-3
 x=-3 => y=-3 ; y'=-1

Tangent:
Passes through $\left(- 3 , - 3\right)$ and has slope $m = - 1$
$\therefore y - \left(- 3\right) = - 1 \left(x - \left(- 3\right)\right)$
$\therefore y + 3 = - x - 3$
$\therefore y = - x - 6$

Normal:
Passes through $\left(- 3 , - 3\right)$ and has slope $m = 1$
$\therefore y - \left(- 3\right) = 1 \left(x - \left(- 3\right)\right)$
$\therefore y + 3 = x + 3$
$\therefore y = x$ 