Graph |y|<x^2-4x+3

$| y | < x^{2} - 4 x + 3$

Step 1

Write $| y | < x^{2} - 4 x + 3$ as a piecewise.

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

$y \geq 0$

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

$y < x^{2} - 4 x + 3$

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

$y < 0$

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

$- y < x^{2} - 4 x + 3$

Write as a piecewise.

${\begin{cases} y < x^{2} - 4 x + 3 & y \geq 0 \\ - y < x^{2} - 4 x + 3 & y < 0 \end{cases}$

Step 2

Find the intersection of $y < x^{2} - 4 x + 3$ and $y \geq 0$ .

$0 \leq y < x^{2} - 4 x + 3$

Step 3

Solve $- y < x^{2} - 4 x + 3$ when $y < 0$ .

Divide each term in $- y < x^{2} - 4 x + 3$ by $- 1$ and simplify.

Divide each term in $- y < x^{2} - 4 x + 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} > \frac{x^{2}}{- 1} + \frac{- 4 x}{- 1} + \frac{3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} > \frac{x^{2}}{- 1} + \frac{- 4 x}{- 1} + \frac{3}{- 1}$

Divide $y$ by $1$ .

$y > \frac{x^{2}}{- 1} + \frac{- 4 x}{- 1} + \frac{3}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$y > - 1 \cdot x^{2} + \frac{- 4 x}{- 1} + \frac{3}{- 1}$

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$y > - x^{2} + \frac{- 4 x}{- 1} + \frac{3}{- 1}$

Move the negative one from the denominator of $\frac{- 4 x}{- 1}$ .

$y > - x^{2} - 1 \cdot (- 4 x) + \frac{3}{- 1}$

Rewrite $- 1 \cdot (- 4 x)$ as $- (- 4 x)$ .

$y > - x^{2} - (- 4 x) + \frac{3}{- 1}$

Multiply $- 4$ by $- 1$ .

$y > - x^{2} + 4 x + \frac{3}{- 1}$

Divide $3$ by $- 1$ .

$y > - x^{2} + 4 x - 3$

Find the intersection of $y > - x^{2} + 4 x - 3$ and $y < 0$ .

$- x^{2} + 4 x - 3 < y < 0$

Step 4

Find the union of the solutions.

$N o (Min i m u m) < y < N o (Max i m u m)$

Step 5

Graph |y|<x^2-4x+3

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