Graph y>-x+2 y>=1/3x-2

$y > - x + 2$ $y \geq \frac{1}{3} x - 2$

Graph $y > - x + 2$ .

Use the slope-intercept form to find the slope and y-intercept.

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

$y = m x + b$

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = - 1$

$b = 2$

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $- 1$

y-intercept: $(0, 2)$

Slope: $- 1$

y-intercept: $(0, 2)$

Graph a dashed line, then shade the area above the boundary line since $y$ is greater than $- x + 2$ .

$y > - x + 2$

Graph $y \geq \frac{1}{3} x - 2$ .

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.

$y = m x + b$

Simplify the right side.

Combine $\frac{1}{3}$ and $x$ .

$y \geq \frac{x}{3} - 2$

Reorder terms.

$y \geq \frac{1}{3} x - 2$

Use the slope-intercept form to find the slope and y-intercept.

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = \frac{1}{3}$

$b = - 2$

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $\frac{1}{3}$

y-intercept: $(0, - 2)$

Slope: $\frac{1}{3}$

y-intercept: $(0, - 2)$

Slope: $\frac{1}{3}$

y-intercept: $(0, - 2)$

Graph a solid line, then shade the area above the boundary line since $y$ is greater than $\frac{1}{3} x - 2$ .

$y \geq \frac{1}{3} x - 2$

Plot each graph on the same coordinate system.

$y > - x + 2$

$y \geq \frac{1}{3} x - 2$