Graph y=|x-3|+1

$y = | x - 3 | + 1$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | x - 3 | + 1$ is $(3, 1)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x - 3$ equal to $0$ . In this case, $x - 3 = 0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Replace the variable $x$ with $3$ in the expression.

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

Simplify $| (3) - 3 | + 1$ .

Simplify each term.

Subtract $3$ from $3$ .

$y = | 0 | + 1$

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

$y = 0 + 1$

Add $0$ and $1$ .

$y = 1$

The absolute value vertex is $(3, 1)$ .

$(3, 1)$

Step 2

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:

${x | x \in ℝ}$

Step 3

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 $1$ into $f (x) = | x - 3 | + 1$ . In this case, the point is $(1, 3)$ .

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

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

Simplify the result.

Simplify each term.

Subtract $3$ from $1$ .

$f (1) = | - 2 | + 1$

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (1) = 2 + 1$

Add $2$ and $1$ .

$f (1) = 3$

The final answer is $3$ .

$y = 3$

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

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

$f (2) = | (2) - 3 | + 1$

Simplify the result.

Simplify each term.

Subtract $3$ from $2$ .

$f (2) = | - 1 | + 1$

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (2) = 1 + 1$

Add $1$ and $1$ .

$f (2) = 2$

The final answer is $2$ .

$y = 2$

Substitute the $x$ value $5$ into $f (x) = | x - 3 | + 1$ . In this case, the point is $(5, 3)$ .

Replace the variable $x$ with $5$ in the expression.

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

Simplify the result.

Simplify each term.

Subtract $3$ from $5$ .

$f (5) = | 2 | + 1$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (5) = 2 + 1$

Add $2$ and $1$ .

$f (5) = 3$

The final answer is $3$ .

$y = 3$

The absolute value can be graphed using the points around the vertex $(3, 1), (1, 3), (2, 2), (4, 2), (5, 3)$

$\begin{array}{c} x & y \\ 1 & 3 \\ 2 & 2 \\ 3 & 1 \\ 4 & 2 \\ 5 & 3 \end{array}$

Step 4

Graph y=|x-3|+1

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