Graph y<1/(x+2)+1

$y < \frac{1}{x + 2} + 1$

Step 1

Simplify $\frac{1}{x + 2} + 1$ .

Write $1$ as a fraction with a common denominator.

$y < \frac{1}{x + 2} + \frac{x + 2}{x + 2}$

Combine the numerators over the common denominator.

$y < \frac{1 + x + 2}{x + 2}$

Add $1$ and $2$ .

$y < \frac{x + 3}{x + 2}$

Step 2

Find the slope and the y-intercept for the boundary line.

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Split the fraction $\frac{x + 3}{x + 2}$ into two fractions.

$y < \frac{x}{x + 2} + \frac{3}{x + 2}$

Rewrite in slope-intercept form.

$y = \frac{1}{1 + 2} x + \frac{3}{x + 2}$

The equation is not linear, so a constant slope does not exist.

Not Linear

Step 3

Graph a dashed line, then shade the area below the boundary line since $y$ is less than $\frac{x + 3}{x + 2}$ .

$y < \frac{x + 3}{x + 2}$

Step 4