Graph f(x) = square root of 9-x^2

$f (x) = \sqrt{9 - x^{2}}$

Step 1

Find the domain for $y = \sqrt{9 - x^{2}}$ 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 $\sqrt{(3 + x) (3 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(3 + x) (3 - x) \geq 0$

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 $0$ .

$3 + x = 0$

$3 - x = 0$

Set $3 + x$ equal to $0$ and solve for $x$ .

Set $3 + x$ equal to $0$ .

$3 + x = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $3 - x$ equal to $0$ and solve for $x$ .

Set $3 - x$ equal to $0$ .

$3 - x = 0$

Solve $3 - x = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

The final solution is all the values that make $(3 + x) (3 - x) \geq 0$ true.

$x = - 3, 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$x > 3$

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 $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$(3 - 6) (3 - (- 6)) \geq 0$

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$(3 + 0) (3 - (0)) \geq 0$

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$(3 + 6) (3 - (6)) \geq 0$

The left side $- 27$ 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.

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 3$

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

Interval Notation:

$[- 3, 3]$

Set-Builder Notation:

${x | - 3 \leq x \leq 3}$

Interval Notation:

$[- 3, 3]$

Set-Builder Notation:

${x | - 3 \leq x \leq 3}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{(3 + x) (3 - x)}$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $3$ from $3$ .

$f (- 3) = \sqrt{0 (3 - (- 3))}$

Multiply $- 1$ by $- 3$ .

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

Add $3$ and $3$ .

$f (- 3) = \sqrt{0 \cdot 6}$

Multiply $0$ by $6$ .

$f (- 3) = \sqrt{0}$

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

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

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

$f (- 3) = 0$

The final answer is $0$ .

$0$

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

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

Simplify the result.

Remove parentheses.

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

Add $3$ and $3$ .

$f (3) = \sqrt{6 (3 - (3))}$

Multiply $- 1$ by $3$ .

$f (3) = \sqrt{6 (3 - 3)}$

Subtract $3$ from $3$ .

$f (3) = \sqrt{6 \cdot 0}$

Multiply $6$ by $0$ .

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

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

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

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

$f (3) = 0$

The final answer is $0$ .

$0$

Step 3

The end points are $(- 3, 0), (3, 0)$ .

$(- 3, 0), (3, 0)$

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

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

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

Simplify the result.

Remove parentheses.

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

Subtract $2$ from $3$ .

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

Multiply $3 - (- 2)$ by $1$ .

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

Multiply $- 1$ by $- 2$ .

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

Add $3$ and $2$ .

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

The final answer is $\sqrt{5}$ .

$y = \sqrt{5}$

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

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

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

Simplify the result.

Remove parentheses.

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

Subtract $1$ from $3$ .

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

Multiply $- 1$ by $- 1$ .

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

Add $3$ and $1$ .

$f (- 1) = \sqrt{2 \cdot 4}$

Multiply $2$ by $4$ .

$f (- 1) = \sqrt{8}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

Rewrite $4$ as $2^{2}$ .

$f (- 1) = \sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

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

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

$y = 2 \sqrt{2}$

Substitute the $x$ value $0$ into $f (x) = \sqrt{(3 + x) (3 - x)}$ . In this case, the point is $(0, 3)$ .

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

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

Simplify the result.

Remove parentheses.

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

Add $3$ and $0$ .

$f (0) = \sqrt{3 (3 - (0))}$

Multiply $- 1$ by $0$ .

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

Add $3$ and $0$ .

$f (0) = \sqrt{3 \cdot 3}$

Multiply $3$ by $3$ .

$f (0) = \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

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

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

$f (0) = 3$

The final answer is $3$ .

$y = 3$

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

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

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

Simplify the result.

Remove parentheses.

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

Add $3$ and $1$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $3$ .

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

Multiply $4$ by $2$ .

$f (1) = \sqrt{8}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

Rewrite $4$ as $2^{2}$ .

$f (1) = \sqrt{2^{2} \cdot 2}$

Pull terms out from under the radical.

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

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

$y = 2 \sqrt{2}$

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

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

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

Simplify the result.

Remove parentheses.

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

Add $3$ and $2$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $3$ .

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

Multiply $5$ by $1$ .

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

The final answer is $\sqrt{5}$ .

$y = \sqrt{5}$

The square root can be graphed using the points around the vertex $(- 3, 0), (3, 0), (- 2, 2.24), (- 1, 2.83), (0, 3), (1, 2.83), (2, 2.24)$

$\begin{array}{c} x & y \\ - 3 & 0 \\ - 2 & 2.24 \\ - 1 & 2.83 \\ 0 & 3 \\ 1 & 2.83 \\ 2 & 2.24 \\ 3 & 0 \end{array}$

Step 5

Graph f(x) = square root of 9-x^2

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