Graph f(x) = square root of x-3+1

$f (x) = \sqrt{x - 3} + 1$

Step 1

Find the domain for $y = \sqrt{x - 3} + 1$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

Set the radicand in $\sqrt{x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${x | x \geq 3}$

Interval Notation:

$[3, \infty)$

Set-Builder Notation:

${x | x \geq 3}$

Step 2

To find the radical expression end point, substitute the $x$ value $3$ , which is the least value in the domain, into $f (x) = \sqrt{x - 3} + 1$ .

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

$f (3) = \sqrt{(3) - 3} + 1$

Simplify the result.

Simplify each term.

Subtract $3$ from $3$ .

$f (3) = \sqrt{0} + 1$

Rewrite $0$ as $0^{2}$ .

$f (3) = \sqrt{0^{2}} + 1$

Pull terms out from under the radical, assuming positive real numbers.

$f (3) = 0 + 1$

Add $0$ and $1$ .

$f (3) = 1$

The final answer is $1$ .

$1$

Step 3

The radical expression end point is $(3, 1)$ .

$(3, 1)$

Step 4

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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

Replace the variable $x$ with $4$ in the expression.

$f (4) = \sqrt{(4) - 3} + 1$

Simplify the result.

Simplify each term.

Subtract $3$ from $4$ .

$f (4) = \sqrt{1} + 1$

Any root of $1$ is $1$ .

$f (4) = 1 + 1$

Add $1$ and $1$ .

$f (4) = 2$

The final answer is $2$ .

$y = 2$

Substitute the $x$ value $5$ into $f (x) = \sqrt{x - 3} + 1$ . In this case, the point is $(5, \sqrt{2} + 1)$ .

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

$f (5) = \sqrt{(5) - 3} + 1$

Simplify the result.

Subtract $3$ from $5$ .

$f (5) = \sqrt{2} + 1$

The final answer is $\sqrt{2} + 1$ .

$y = \sqrt{2} + 1$

The square root can be graphed using the points around the vertex $(3, 1), (4, 2), (5, 2.41)$

$\begin{array}{c} x & y \\ 3 & 1 \\ 4 & 2 \\ 5 & 2.41 \end{array}$

Step 5

Graph f(x) = square root of x-3+1

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