Graph S(t)=1/2 square root of x-3+1

$S (t) = \frac{1}{2} \sqrt{x - 3} + 1$

Find the domain for $y = \frac{1}{2} \cdot \sqrt{x - 3} + 1$ 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:

${x | x \geq 3}$

To find the radical expression end point, substitute the $x$ value $3$ , which is the least value in the domain, into $f (x) = \frac{\sqrt{x - 3} + 2}{2}$ .

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

$f (3) = \frac{\sqrt{(3) - 3} + 2}{2}$

Simplify the result.

Simplify the numerator.

Subtract $3$ from $3$ .

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

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

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

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

$f (3) = \frac{0 + 2}{2}$

Add $0$ and $2$ .

$f (3) = \frac{2}{2}$

Divide $2$ by $2$ .

$f (3) = 1$

The final answer is $1$ .

$1$

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

$(3, 1)$

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

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

$f (4) = \frac{\sqrt{(4) - 3} + 2}{2}$

Simplify the result.

Simplify the numerator.

Subtract $3$ from $4$ .

$f (4) = \frac{\sqrt{1} + 2}{2}$

Any root of $1$ is $1$ .

$f (4) = \frac{1 + 2}{2}$

Add $1$ and $2$ .

$f (4) = \frac{3}{2}$

The final answer is $\frac{3}{2}$ .

$y = \frac{3}{2}$

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

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

$f (5) = \frac{\sqrt{(5) - 3} + 2}{2}$

Simplify the result.

Subtract $3$ from $5$ .

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

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

$y = \frac{\sqrt{2} + 2}{2}$

The square root can be graphed using the points around the vertex $(3, 1), (4, 1.5), (5, 1.71)$

$\begin{array}{c} x & y \\ 3 & 1 \\ 4 & 1.5 \\ 5 & 1.71 \end{array}$

Graph S(t)=1/2 square root of x-3+1

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