Graph y = square root of 2x-x^2

$y = \sqrt{2 x - x^{2}}$

Step 1

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

$x (2 - x) \geq 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$2 - x = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 - x$ equal to $0$ and solve for $x$ .

Set $2 - x$ equal to $0$ .

$2 - x = 0$

Solve $2 - x = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

The final solution is all the values that make $x (2 - x) \geq 0$ true.

$x = 0, 2$

Use each root to create test intervals.

$x < 0$

$0 < x < 2$

$x > 2$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$(- 2) (2 - (- 2)) \geq 0$

The left side $- 8$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$(1) (2 - (1)) \geq 0$

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$(4) (2 - (4)) \geq 0$

The left side $- 8$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 2$ True

$x > 2$ False

$x < 0$ False

$0 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$0 \leq x \leq 2$

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

Interval Notation:

$[0, 2]$

Set-Builder Notation:

${x | 0 \leq x \leq 2}$

Interval Notation:

$[0, 2]$

Set-Builder Notation:

${x | 0 \leq x \leq 2}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{x (2 - x)}$ .

Replace the variable $x$ with $0$ in the expression.

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

Simplify the result.

Multiply $0$ by $2 - (0)$ .

$f (0) = \sqrt{0}$

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

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

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

$f (0) = 0$

The final answer is $0$ .

$0$

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

$f (2) = \sqrt{(2) (2 - (2))}$

Simplify the result.

Multiply $- 1$ by $2$ .

$f (2) = \sqrt{2 (2 - 2)}$

Subtract $2$ from $2$ .

$f (2) = \sqrt{2 \cdot 0}$

Multiply $2$ by $0$ .

$f (2) = \sqrt{0}$

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

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

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

$f (2) = 0$

The final answer is $0$ .

$0$

Step 3

The end points are $(0, 0), (2, 0)$ .

$(0, 0), (2, 0)$

Step 4

The square root can be graphed using the points around the vertex $(0, 0), (2, 0),$

$\begin{array}{c} x & y \\ 0 & 0 \\ 2 & 0 \end{array}$

Step 5

Graph y = square root of 2x-x^2

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