Graph y=1+5/2 square root of x-1

$y = 1 + \frac{5}{2} \sqrt{x - 1}$

Find the domain for $y = 1 + \frac{5}{2} \cdot \sqrt{x - 1}$ 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 $\sqrt{x - 1}$ greater than or equal to $0$ to find where the expression is defined.

$x - 1 \geq 0$

Add $1$ to both sides of the inequality.

$x \geq 1$

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

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${x | x \geq 1}$

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${x | x \geq 1}$

To find the radical expression end point, substitute the $x$ value $1$ , which is the least value in the domain, into $f (x) = \frac{2 + 5 \sqrt{x - 1}}{2}$ .

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

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

Simplify the result.

Simplify the numerator.

Subtract $1$ from $1$ .

$f (1) = \frac{2 + 5 \sqrt{0}}{2}$

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

$f (1) = \frac{2 + 5 \sqrt{0^{2}}}{2}$

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

$f (1) = \frac{2 + 5 \cdot 0}{2}$

Multiply $5$ by $0$ .

$f (1) = \frac{2 + 0}{2}$

Add $2$ and $0$ .

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

Divide $2$ by $2$ .

$f (1) = 1$

The final answer is $1$ .

$1$

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

$(1, 1)$

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 $2$ into $f (x) = \frac{2 + 5 \sqrt{x - 1}}{2}$ . In this case, the point is $(2, \frac{7}{2})$ .

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

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

Simplify the result.

Simplify the numerator.

Subtract $1$ from $2$ .

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

Any root of $1$ is $1$ .

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

Multiply $5$ by $1$ .

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

Add $2$ and $5$ .

$f (2) = \frac{7}{2}$

The final answer is $\frac{7}{2}$ .

$y = \frac{7}{2}$

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

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

$f (3) = \frac{2 + 5 \sqrt{(3) - 1}}{2}$

Simplify the result.

Subtract $1$ from $3$ .

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

The final answer is $\frac{2 + 5 \sqrt{2}}{2}$ .

$y = \frac{2 + 5 \sqrt{2}}{2}$

The square root can be graphed using the points around the vertex $(1, 1), (2, 3.5), (3, 4.54)$

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 3.5 \\ 3 & 4.54 \end{array}$

Graph y=1+5/2 square root of x-1

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