# Find the equation of the tangent to the curve x + xy + y = 5 at x = 5 help?

Jan 10, 2018

$y = \frac{5}{6} - \frac{1}{6} x$

#### Explanation:

Start by finding the y-value:

$5 + 5 y + y = 5$

$6 y = 0$

$y = 0$

Now we find the derivative using implicit differentiation.

$1 + y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

$x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} = - 1 - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x + 1\right) = - 1 - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1 - y}{x + 1}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y + 1}{x + 1}$

At $\left(5 , 0\right)$, the derivative will have value

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{6}$

Now use point-slope form to find the equation:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 0 = - \frac{1}{6} \left(x - 5\right)$

$y = - \frac{1}{6} x + \frac{5}{6}$

Hopefully this helps!

Jan 10, 2018

$y = - \frac{1}{6} x + \frac{5}{6}$

#### Explanation:

Begin by finding the y coordinate at $x = 5$:

$5 + 5 y + y = 5$

$5 y + y = 0$

$6 y = 0$

$y = 0$

The point of tangency is $\left(5 , 0\right)$

Compute the first derivative of the curve:

$1 + y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\left(x + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(y + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y + 1}{x + 1}$

The slope, m, of the tangent line is the first derivative evaluated at the point $\left(5 , 0\right)$:

$m = - \frac{0 + 1}{5 + 1}$

$m = - \frac{1}{6}$

Use the point-slope form for the equation of a line:

$y = m \left(x - {x}_{0}\right) + {y}_{0}$

$y = - \frac{1}{6} \left(x - 5\right) + 0$

$y = - \frac{1}{6} x + \frac{5}{6}$

Here is the a graph of the curve and the tangent line: