Algebra Examples

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

Step 1

Rewrite the equation as $x^{2} - 4 x + 3 = | y |$ .

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

Step 2

Find the absolute value vertex. In this case, the vertex for $x^{2} - 4 x + 3 = | y |$ is $(0, 0)$ .

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Step 2.1

To find the $y$ coordinate of the vertex, set the inside of the absolute value $y$ equal to $0$ . In this case, $y = 0$ .

$y = 0$

Step 2.2

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

$x = | 0 |$

Step 2.3

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

$x = 0$

Step 2.4

The absolute value vertex is $(0, 0)$ .

$(0, 0)$

Step 3

The domain is the set of all valid $x$ values. Use the graph to find the domain.

$[0, \infty)$

${x | x \geq 0}$

Step 4

For each $x$ value, there is one positive $y$ value and one negative $y$ 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.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

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

Step 4.1.2.1.2

Multiply $- 4$ by $- 2$ .

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

Step 4.1.2.2

Simplify by adding numbers.

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Step 4.1.2.2.1

Add $4$ and $8$ .

$f (- 2) = 12 + 3$

Step 4.1.2.2.2

Add $12$ and $3$ .

$f (- 2) = 15$

Step 4.1.2.3

The final answer is $15$ .

$y = 15$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

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

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

Step 4.2.2.1.2

Multiply $- 4$ by $- 2$ .

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

Step 4.2.2.2

Simplify the expression.

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Step 4.2.2.2.1

Add $4$ and $8$ .

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

Step 4.2.2.2.2

Add $12$ and $3$ .

$f (- 2) = - 1 \cdot 15$

Step 4.2.2.2.3

Multiply $- 1$ by $15$ .

$f (- 2) = - 15$

Step 4.2.2.3

The final answer is $- 15$ .

$y = - 15$

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

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

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

Step 4.3.2.1.2

Multiply $- 4$ by $- 1$ .

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

Step 4.3.2.2

Simplify by adding numbers.

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Step 4.3.2.2.1

Add $1$ and $4$ .

$f (- 1) = 5 + 3$

Step 4.3.2.2.2

Add $5$ and $3$ .

$f (- 1) = 8$

Step 4.3.2.3

The final answer is $8$ .

$y = 8$

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify the result.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

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

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

Step 4.4.2.1.2

Multiply $- 4$ by $- 1$ .

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

Step 4.4.2.2

Simplify the expression.

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Step 4.4.2.2.1

Add $1$ and $4$ .

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

Step 4.4.2.2.2

Add $5$ and $3$ .

$f (- 1) = - 1 \cdot 8$

Step 4.4.2.2.3

Multiply $- 1$ by $8$ .

$f (- 1) = - 8$

Step 4.4.2.3

The final answer is $- 8$ .

$y = - 8$

Step 4.5

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

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

Step 5