Graph y+4x<=x^2+2

$y + 4 x \leq x^{2} + 2$

Subtract $4 x$ from both sides of the inequality.

$y \leq x^{2} + 2 - 4 x$

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.

$y = m x + b$

Move $2$ .

$y \leq x^{2} - 4 x + 2$

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

Not Linear

Graph a solid line, then shade the area below the boundary line since $y$ is less than $x^{2} + 2 - 4 x$ .

$y \leq x^{2} + 2 - 4 x$