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

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

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

Substitute the $x$ value $- 2$ into $f (x) = | x^{2} - 4 x + 3 |$ . In this case, the point is $(- 2, 15)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = | {(- 2)}^{2} - 4 \cdot - 2 + 3 |$

Simplify the result.

Simplify each term.

Raise $- 2$ to the power of $2$ .

$f (- 2) = | 4 - 4 \cdot - 2 + 3 |$

Multiply $- 4$ by $- 2$ .

$f (- 2) = | 4 + 8 + 3 |$

Simplify by adding numbers.

Add $4$ and $8$ .

$f (- 2) = | 12 + 3 |$

Add $12$ and $3$ .

$f (- 2) = | 15 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $15$ is $15$ .

$f (- 2) = 15$

The final answer is $15$ .

$y = 15$

Substitute the $x$ value $- 1$ into $f (x) = | x^{2} - 4 x + 3 |$ . In this case, the point is $(- 1, 8)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = | {(- 1)}^{2} - 4 \cdot - 1 + 3 |$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = | 1 - 4 \cdot - 1 + 3 |$

Multiply $- 4$ by $- 1$ .

$f (- 1) = | 1 + 4 + 3 |$

Simplify by adding numbers.

Add $1$ and $4$ .

$f (- 1) = | 5 + 3 |$

Add $5$ and $3$ .

$f (- 1) = | 8 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$f (- 1) = 8$

The final answer is $8$ .

$y = 8$

Substitute the $x$ value $0$ into $f (x) = | x^{2} - 4 x + 3 |$ . In this case, the point is $(0, 3)$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = | {(0)}^{2} - 4 \cdot 0 + 3 |$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$f (0) = | 0 - 4 \cdot 0 + 3 |$

Multiply $- 4$ by $0$ .

$f (0) = | 0 + 0 + 3 |$

Simplify by adding numbers.

Add $0$ and $0$ .

$f (0) = | 0 + 3 |$

Add $0$ and $3$ .

$f (0) = | 3 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$f (0) = 3$

The final answer is $3$ .

$y = 3$

Substitute the $x$ value $1$ into $f (x) = | x^{2} - 4 x + 3 |$ . In this case, the point is $(1, 0)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = | {(1)}^{2} - 4 \cdot 1 + 3 |$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = | 1 - 4 \cdot 1 + 3 |$

Multiply $- 4$ by $1$ .

$f (1) = | 1 - 4 + 3 |$

Simplify by adding and subtracting.

Subtract $4$ from $1$ .

$f (1) = | - 3 + 3 |$

Add $- 3$ and $3$ .

$f (1) = | 0 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$f (1) = 0$

The final answer is $0$ .

$y = 0$

The absolute value can be graphed using the points around the vertex $(- 2, 15), (- 1, 8), (0, 3), (1, 0)$

$\begin{array}{c} x & y \\ - 2 & 15 \\ - 1 & 8 \\ 0 & 3 \\ 1 & 0 \end{array}$

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

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