Algebra Examples

y = \sqrt{1 - x} + 2

$y = \sqrt{1 - x} + 2$

Step 1

Reorder

1

$1$ and

- x

$- x$ .

y = \sqrt{- x + 1} + 2

$y = \sqrt{- x + 1} + 2$

Step 2

Find the domain for

y = \sqrt{1 - x} + 2

$y = \sqrt{1 - x} + 2$ so that a list of

x

$x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 2.1

Set the radicand in

\sqrt{- x + 1}

$\sqrt{- x + 1}$ greater than or equal to

0

$0$ to find where the expression is defined.

- x + 1 \geq 0

$- x + 1 \geq 0$

Step 2.2

Solve for

x

$x$ .

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Step 2.2.1

Subtract

1

$1$ from both sides of the inequality.

- x \geq - 1

$- x \geq - 1$

Step 2.2.2

Divide each term in

- x \geq - 1

$- x \geq - 1$ by

- 1

$- 1$ and simplify.

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Step 2.2.2.1

Divide each term in

- x \geq - 1

$- x \geq - 1$ by

- 1

$- 1$ . 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{- 1}{- 1}

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

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Dividing two negative values results in a positive value.

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

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

Step 2.2.2.2.2

Divide

x

$x$ by

1

$1$ .

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

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

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

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

Step 2.2.2.3

Simplify the right side.

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Step 2.2.2.3.1

Divide

- 1

$- 1$ by

- 1

$- 1$ .

x \leq 1

$x \leq 1$

x \leq 1

$x \leq 1$

x \leq 1

$x \leq 1$

x \leq 1

$x \leq 1$

Step 2.3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

(- \infty, 1]

$(- \infty, 1]$

Set-Builder Notation:

{x | x \leq 1}

${x | x \leq 1}$

Interval Notation:

(- \infty, 1]

$(- \infty, 1]$

Set-Builder Notation:

{x | x \leq 1}

${x | x \leq 1}$

Step 3

To find the radical expression end point, substitute the

x

$x$ value

1

$1$ , which is the least value in the domain, into

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

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

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Step 3.1

Replace the variable

x

$x$ with

1

$1$ in the expression.

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

Step 3.2.1.2

Add

- 1

$- 1$ and

1

$1$ .

f (1) = \sqrt{0} + 2

$f (1) = \sqrt{0} + 2$

Step 3.2.1.3

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

f (1) = \sqrt{0^{2}} + 2

$f (1) = \sqrt{0^{2}} + 2$

Step 3.2.1.4

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

f (1) = 0 + 2

$f (1) = 0 + 2$

f (1) = 0 + 2

$f (1) = 0 + 2$

Step 3.2.2

Add

0

$0$ and

2

$2$ .

f (1) = 2

$f (1) = 2$

Step 3.2.3

The final answer is

2

$2$ .

2

$2$

2

$2$

2

$2$

Step 4

The radical expression end point is

(1, 2)

$(1, 2)$ .

(1, 2)

$(1, 2)$

Step 5

Select a few

x

$x$ values from the domain. It would be more useful to select the values so that they are next to the

x

$x$ value of the radical expression end point.

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Step 5.1

Substitute the

x

$x$ value

- 1

$- 1$ into

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

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

(- 1, \sqrt{2} + 2)

$(- 1, \sqrt{2} + 2)$ .

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Step 5.1.1

Replace the variable

x

$x$ with

- 1

$- 1$ in the expression.

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

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

Step 5.1.2

Simplify the result.

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Step 5.1.2.1

Simplify each term.

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Step 5.1.2.1.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

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

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

Step 5.1.2.1.2

Add

1

$1$ and

1

$1$ .

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

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

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

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

Step 5.1.2.2

The final answer is

\sqrt{2} + 2

$\sqrt{2} + 2$ .

y = \sqrt{2} + 2

$y = \sqrt{2} + 2$

y = \sqrt{2} + 2

$y = \sqrt{2} + 2$

y = \sqrt{2} + 2

$y = \sqrt{2} + 2$

Step 5.2

Substitute the

x

$x$ value

0

$0$ into

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

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

(0, 3)

$(0, 3)$ .

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Step 5.2.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = \sqrt{- (0) + 1} + 2

$f (0) = \sqrt{- (0) + 1} + 2$

Step 5.2.2

Simplify the result.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply

- 1

$- 1$ by

0

$0$ .

f (0) = \sqrt{0 + 1} + 2

$f (0) = \sqrt{0 + 1} + 2$

Step 5.2.2.1.2

Add

0

$0$ and

1

$1$ .

f (0) = \sqrt{1} + 2

$f (0) = \sqrt{1} + 2$

Step 5.2.2.1.3

Any root of

1

$1$ is

1

$1$ .

f (0) = 1 + 2

$f (0) = 1 + 2$

f (0) = 1 + 2

$f (0) = 1 + 2$

Step 5.2.2.2

Add

1

$1$ and

2

$2$ .

f (0) = 3

$f (0) = 3$

Step 5.2.2.3

The final answer is

3

$3$ .

y = 3

$y = 3$

y = 3

$y = 3$

y = 3

$y = 3$

Step 5.3

The square root can be graphed using the points around the vertex

(1, 2), (- 1, 3.41), (0, 3)

$(1, 2), (- 1, 3.41), (0, 3)$

\begin{array}{c} x & y \\ - 1 & 3.41 \\ 0 & 3 \\ 1 & 2 \end{array}

$\begin{array}{c} x & y \\ - 1 & 3.41 \\ 0 & 3 \\ 1 & 2 \end{array}$

\begin{array}{c} x & y \\ - 1 & 3.41 \\ 0 & 3 \\ 1 & 2 \end{array}

$\begin{array}{c} x & y \\ - 1 & 3.41 \\ 0 & 3 \\ 1 & 2 \end{array}$

Step 6