Graph f(x)=5/2 square root of -(x-6)

$f (x) = \frac{5}{2} \sqrt{- (x - 6)}$

Find the domain for $y = \frac{5}{2} \cdot \sqrt{- (x - 6)}$ 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 $\sqrt{- (x - 6)}$ greater than or equal to $0$ to find where the expression is defined.

$- (x - 6) \geq 0$

Solve for $x$ .

Simplify $- (x - 6)$ .

Apply the distributive property.

$- x - - 6 \geq 0$

Multiply $- 1$ by $- 6$ .

$- x + 6 \geq 0$

Subtract $6$ from both sides of the inequality.

$- x \geq - 6$

Divide each term in $- x \geq - 6$ by $- 1$ and simplify.

Divide each term in $- x \geq - 6$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x \leq 6$

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

Interval Notation:

$(- \infty, 6]$

Set-Builder Notation:

${x | x \leq 6}$

Interval Notation:

$(- \infty, 6]$

Set-Builder Notation:

${x | x \leq 6}$

To find the radical expression end point, substitute the $x$ value $6$ , which is the least value in the domain, into $f (x) = \frac{5 \sqrt{- (x - 6)}}{2}$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = \frac{5 \sqrt{- ((6) - 6)}}{2}$

Simplify the result.

Simplify the numerator.

Subtract $6$ from $6$ .

$f (6) = \frac{5 \sqrt{- 0}}{2}$

Multiply $- 1$ by $0$ .

$f (6) = \frac{5 \sqrt{0}}{2}$

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

$f (6) = \frac{5 \sqrt{0^{2}}}{2}$

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

$f (6) = \frac{5 \cdot 0}{2}$

Simplify the expression.

Multiply $5$ by $0$ .

$f (6) = \frac{0}{2}$

Divide $0$ by $2$ .

$f (6) = 0$

The final answer is $0$ .

$0$

The radical expression end point is $(6, 0)$ .

$(6, 0)$

The square root can be graphed using the points around the vertex $(6, 0),$

$\begin{array}{c} x & y \\ 6 & 0 \end{array}$

Graph f(x)=5/2 square root of -(x-6)

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