Graph y = square root of 16-x^2

$y = \sqrt{16 - x^{2}}$

Step 1

Find the domain for $y = \sqrt{16 - x^{2}}$ 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 $\sqrt{(4 + x) (4 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(4 + x) (4 - x) \geq 0$

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 $0$ .

$4 + x = 0$

$4 - x = 0$

Set $4 + x$ equal to $0$ and solve for $x$ .

Set $4 + x$ equal to $0$ .

$4 + x = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $4 - x$ equal to $0$ and solve for $x$ .

Set $4 - x$ equal to $0$ .

$4 - x = 0$

Solve $4 - x = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x = - 4$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

The final solution is all the values that make $(4 + x) (4 - x) \geq 0$ true.

$x = - 4, 4$

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 4$

$x > 4$

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 $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$(4 - 6) (4 - (- 6)) \geq 0$

The left side $- 20$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $16$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$(4 + 6) (4 - (6)) \geq 0$

The left side $- 20$ 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.

$x < - 4$ False

$- 4 < x < 4$ True

$x > 4$ False

$x < - 4$ False

$- 4 < x < 4$ True

$x > 4$ False

The solution consists of all of the true intervals.

$- 4 \leq x \leq 4$

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

Interval Notation:

$[- 4, 4]$

Set-Builder Notation:

${x | - 4 \leq x \leq 4}$

Interval Notation:

$[- 4, 4]$

Set-Builder Notation:

${x | - 4 \leq x \leq 4}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{(4 + x) (4 - x)}$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $4$ from $4$ .

$f (- 4) = \sqrt{0 (4 - (- 4))}$

Multiply $- 1$ by $- 4$ .

$f (- 4) = \sqrt{0 (4 + 4)}$

Add $4$ and $4$ .

$f (- 4) = \sqrt{0 \cdot 8}$

Multiply $0$ by $8$ .

$f (- 4) = \sqrt{0}$

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

$f (- 4) = \sqrt{0^{2}}$

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

$f (- 4) = 0$

The final answer is $0$ .

$0$

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

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

Simplify the result.

Remove parentheses.

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

Add $4$ and $4$ .

$f (4) = \sqrt{8 (4 - (4))}$

Multiply $- 1$ by $4$ .

$f (4) = \sqrt{8 (4 - 4)}$

Subtract $4$ from $4$ .

$f (4) = \sqrt{8 \cdot 0}$

Multiply $8$ by $0$ .

$f (4) = \sqrt{0}$

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

$f (4) = \sqrt{0^{2}}$

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

$f (4) = 0$

The final answer is $0$ .

$0$

Step 3

The end points are $(- 4, 0), (4, 0)$ .

$(- 4, 0), (4, 0)$

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 $- 3$ into $f (x) = \sqrt{(4 + x) (4 - x)}$ . In this case, the point is $(- 3, \sqrt{7})$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $3$ from $4$ .

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

Multiply $4 - (- 3)$ by $1$ .

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

Multiply $- 1$ by $- 3$ .

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

Add $4$ and $3$ .

$f (- 3) = \sqrt{7}$

The final answer is $\sqrt{7}$ .

$y = \sqrt{7}$

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

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

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

Simplify the result.

Remove parentheses.

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

Subtract $2$ from $4$ .

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

Multiply $- 1$ by $- 2$ .

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

Add $4$ and $2$ .

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

Multiply $2$ by $6$ .

$f (- 2) = \sqrt{12}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

$y = 2 \sqrt{3}$

Substitute the $x$ value $- 1$ into $f (x) = \sqrt{(4 + x) (4 - x)}$ . In this case, the point is $(- 1, \sqrt{15})$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $1$ from $4$ .

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

Multiply $- 1$ by $- 1$ .

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

Add $4$ and $1$ .

$f (- 1) = \sqrt{3 \cdot 5}$

Multiply $3$ by $5$ .

$f (- 1) = \sqrt{15}$

The final answer is $\sqrt{15}$ .

$y = \sqrt{15}$

Substitute the $x$ value $0$ into $f (x) = \sqrt{(4 + x) (4 - x)}$ . In this case, the point is $(0, 4)$ .

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

$f (0) = \sqrt{(4 + 0) (4 - (0))}$

Simplify the result.

Remove parentheses.

$f (0) = \sqrt{(4 + 0) (4 - (0))}$

Add $4$ and $0$ .

$f (0) = \sqrt{4 (4 - (0))}$

Multiply $- 1$ by $0$ .

$f (0) = \sqrt{4 (4 + 0)}$

Add $4$ and $0$ .

$f (0) = \sqrt{4 \cdot 4}$

Multiply $4$ by $4$ .

$f (0) = \sqrt{16}$

Rewrite $16$ as $4^{2}$ .

$f (0) = \sqrt{4^{2}}$

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

$f (0) = 4$

The final answer is $4$ .

$y = 4$

Substitute the $x$ value $1$ into $f (x) = \sqrt{(4 + x) (4 - x)}$ . In this case, the point is $(1, \sqrt{15})$ .

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

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

Simplify the result.

Remove parentheses.

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

Add $4$ and $1$ .

$f (1) = \sqrt{5 (4 - (1))}$

Multiply $- 1$ by $1$ .

$f (1) = \sqrt{5 (4 - 1)}$

Subtract $1$ from $4$ .

$f (1) = \sqrt{5 \cdot 3}$

Multiply $5$ by $3$ .

$f (1) = \sqrt{15}$

The final answer is $\sqrt{15}$ .

$y = \sqrt{15}$

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

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

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

Simplify the result.

Remove parentheses.

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

Add $4$ and $2$ .

$f (2) = \sqrt{6 (4 - (2))}$

Multiply $- 1$ by $2$ .

$f (2) = \sqrt{6 (4 - 2)}$

Subtract $2$ from $4$ .

$f (2) = \sqrt{6 \cdot 2}$

Multiply $6$ by $2$ .

$f (2) = \sqrt{12}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

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

$y = 2 \sqrt{3}$

Substitute the $x$ value $3$ into $f (x) = \sqrt{(4 + x) (4 - x)}$ . In this case, the point is $(3, \sqrt{7})$ .

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

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

Simplify the result.

Remove parentheses.

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

Add $4$ and $3$ .

$f (3) = \sqrt{7 (4 - (3))}$

Multiply $- 1$ by $3$ .

$f (3) = \sqrt{7 (4 - 3)}$

Subtract $3$ from $4$ .

$f (3) = \sqrt{7 \cdot 1}$

Multiply $7$ by $1$ .

$f (3) = \sqrt{7}$

The final answer is $\sqrt{7}$ .

$y = \sqrt{7}$

The square root can be graphed using the points around the vertex $(- 4, 0), (4, 0), (- 3, 2.65), (- 2, 3.46), (- 1, 3.87), (0, 4), (1, 3.87), (2, 3.46), (3, 2.65)$

$\begin{array}{c} x & y \\ - 4 & 0 \\ - 3 & 2.65 \\ - 2 & 3.46 \\ - 1 & 3.87 \\ 0 & 4 \\ 1 & 3.87 \\ 2 & 3.46 \\ 3 & 2.65 \\ 4 & 0 \end{array}$

Step 5

Graph y = square root of 16-x^2

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