Graph y = square root of 25-x^2

$y = \sqrt{25 - x^{2}}$

Step 1

Find the domain for $y = \sqrt{25 - x^{2}}$ 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 $\sqrt{(5 + x) (5 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(5 + x) (5 - x) \geq 0$

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 $0$ .

$5 + x = 0$

$5 - x = 0$

Set $5 + x$ equal to $0$ and solve for $x$ .

Set $5 + x$ equal to $0$ .

$5 + x = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $5 - x$ equal to $0$ and solve for $x$ .

Set $5 - x$ equal to $0$ .

$5 - x = 0$

Solve $5 - x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

The final solution is all the values that make $(5 + x) (5 - x) \geq 0$ true.

$x = - 5, 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 5$

$x > 5$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$(5 - 8) (5 - (- 8)) \geq 0$

The left side $- 39$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

$(5 + 0) (5 - (0)) \geq 0$

The left side $25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$(5 + 8) (5 - (8)) \geq 0$

The left side $- 39$ 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.

$x < - 5$ False

$- 5 < x < 5$ True

$x > 5$ False

$x < - 5$ False

$- 5 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$- 5 \leq x \leq 5$

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

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${x | - 5 \leq x \leq 5}$

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${x | - 5 \leq x \leq 5}$

Step 2

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \sqrt{(5 + x) (5 - x)}$ .

Replace the variable $x$ with $- 5$ in the expression.

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

Simplify the result.

Remove parentheses.

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

Subtract $5$ from $5$ .

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

Multiply $- 1$ by $- 5$ .

$f (- 5) = \sqrt{0 (5 + 5)}$

Add $5$ and $5$ .

$f (- 5) = \sqrt{0 \cdot 10}$

Multiply $0$ by $10$ .

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

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

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

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

$f (- 5) = 0$

The final answer is $0$ .

$0$

Replace the variable $x$ with $5$ in the expression.

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

Simplify the result.

Remove parentheses.

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

Add $5$ and $5$ .

$f (5) = \sqrt{10 (5 - (5))}$

Multiply $- 1$ by $5$ .

$f (5) = \sqrt{10 (5 - 5)}$

Subtract $5$ from $5$ .

$f (5) = \sqrt{10 \cdot 0}$

Multiply $10$ by $0$ .

$f (5) = \sqrt{0}$

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

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

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

$f (5) = 0$

The final answer is $0$ .

$0$

Step 3

The end points are $(- 5, 0), (5, 0)$ .

$(- 5, 0), (5, 0)$

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 $f (x) = \sqrt{(5 + x) (5 - x)}$ . In this case, the point is $(- 3, 4)$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $3$ from $5$ .

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

Multiply $- 1$ by $- 3$ .

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

Add $5$ and $3$ .

$f (- 3) = \sqrt{2 \cdot 8}$

Multiply $2$ by $8$ .

$f (- 3) = \sqrt{16}$

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

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

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

$f (- 3) = 4$

The final answer is $4$ .

$y = 4$

Substitute the $x$ value $- 1$ into $f (x) = \sqrt{(5 + x) (5 - x)}$ . In this case, the point is $(- 1, 2 \sqrt{6})$ .

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

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

Simplify the result.

Remove parentheses.

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

Subtract $1$ from $5$ .

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

Multiply $- 1$ by $- 1$ .

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

Add $5$ and $1$ .

$f (- 1) = \sqrt{4 \cdot 6}$

Multiply $4$ by $6$ .

$f (- 1) = \sqrt{24}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$f (- 1) = \sqrt{4 (6)}$

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

$f (- 1) = \sqrt{2^{2} \cdot 6}$

Pull terms out from under the radical.

$f (- 1) = 2 \sqrt{6}$

The final answer is $2 \sqrt{6}$ .

$y = 2 \sqrt{6}$

Substitute the $x$ value $3$ into $f (x) = \sqrt{(5 + x) (5 - x)}$ . In this case, the point is $(3, 4)$ .

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

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

Simplify the result.

Remove parentheses.

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

Add $5$ and $3$ .

$f (3) = \sqrt{8 (5 - (3))}$

Multiply $- 1$ by $3$ .

$f (3) = \sqrt{8 (5 - 3)}$

Subtract $3$ from $5$ .

$f (3) = \sqrt{8 \cdot 2}$

Multiply $8$ by $2$ .

$f (3) = \sqrt{16}$

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

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

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

$f (3) = 4$

The final answer is $4$ .

$y = 4$

The square root can be graphed using the points around the vertex $(- 5, 0), (5, 0), (- 3, 4), (- 1, 4.9), (1, 4.9), (3, 4)$

$\begin{array}{c} x & y \\ - 5 & 0 \\ - 3 & 4 \\ - 1 & 4.9 \\ 1 & 4.9 \\ 3 & 4 \\ 5 & 0 \end{array}$

Step 5

Graph y = square root of 25-x^2

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