Graph f(x)>-|x-2|+4

$f (x) > - | x - 2 | + 4$

Subtract $f (x)$ from both sides of the inequality.

$0 > - | x - 2 | + 4 - f (x)$

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.

$y = m x + b$

Rewrite so $x$ is on the left side of the inequality.

$- | x - 2 | + 4 - f (x) < 0$

Write $- | x - 2 | + 4 - f (x) < 0$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x - 2 \geq 0$

Add $2$ to both sides of the inequality.

$x \geq 2$

In the piece where $x - 2$ is non-negative, remove the absolute value.

$- (x - 2) + 4 - f (x) < 0$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x - 2 < 0$

Add $2$ to both sides of the inequality.

$x < 2$

In the piece where $x - 2$ is negative, remove the absolute value and multiply by $- 1$ .

$- - (x - 2) + 4 - f (x) < 0$

Write as a piecewise.

${\begin{cases} - (x - 2) + 4 - f (x) < 0 & x \geq 2 \\ - - (x - 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Simplify $- (x - 2) + 4 - f (x) < 0$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - x - - 2 + 4 - f (x) < 0 & x \geq 2 \\ - - (x - 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Multiply $- 1$ by $- 2$ .

${\begin{cases} - x + 2 + 4 - f (x) < 0 & x \geq 2 \\ - - (x - 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Add $2$ and $4$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ - - (x - 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Simplify $- - (x - 2) + 4 - f (x) < 0$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ - (- x - - 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Multiply $- 1$ by $- 2$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ - (- x + 2) + 4 - f (x) < 0 & x < 2 \end{cases}$

Apply the distributive property.

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ - - x - 1 \cdot 2 + 4 - f (x) < 0 & x < 2 \end{cases}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ 1 x - 1 \cdot 2 + 4 - f (x) < 0 & x < 2 \end{cases}$

Multiply $x$ by $1$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ x - 1 \cdot 2 + 4 - f (x) < 0 & x < 2 \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ x - 2 + 4 - f (x) < 0 & x < 2 \end{cases}$

Add $- 2$ and $4$ .

${\begin{cases} - x + 6 - f (x) < 0 & x \geq 2 \\ x + 2 - f (x) < 0 & x < 2 \end{cases}$

Solve $- x + 6 - f (x) < 0$ when $x \geq 2$ .

Solve $- x + 6 - f (x) < 0$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $6$ from both sides of the inequality.

$- x - f (x) < - 6$

Add $f (x)$ to both sides of the inequality.

$- x < - 6 + f (x)$

Divide each term in $- x < - 6 + f (x)$ by $- 1$ and simplify.

Divide each term in $- x < - 6 + f (x)$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{- 6}{- 1} + \frac{f (x)}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{- 6}{- 1} + \frac{f (x)}{- 1}$

Divide $x$ by $1$ .

$x > \frac{- 6}{- 1} + \frac{f (x)}{- 1}$

Simplify the right side.

Simplify each term.

Divide $- 6$ by $- 1$ .

$x > 6 + \frac{f (x)}{- 1}$

Move the negative one from the denominator of $\frac{f (x)}{- 1}$ .

$x > 6 - 1 \cdot f (x)$

Rewrite $- 1 \cdot f (x)$ as $- f (x)$ .

$x > 6 - f (x)$

Find the intersection of $x > 6 - f (x)$ and $x \geq 2$ .

$x > 6 - f (x)$ and $x \geq 2$

Solve $x + 2 - f (x) < 0$ when $x < 2$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$x - f (x) < - 2$

Add $f (x)$ to both sides of the inequality.

$x < - 2 + f (x)$

Find the intersection of $x < - 2 + f (x)$ and $x < 2$ .

$x < N o (Min i m u m)$

Find the union of the solutions.

$x > N o Max i m u m$ or $x < N o Min i m u m$

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

Not Linear

Graph a dashed line, then shade the area above the boundary line since $y$ is greater than $- | x - 2 | + 4 - f (x)$ .

$0 > - | x - 2 | + 4 - f (x)$

Graph f(x)>-|x-2|+4

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