Precalculus Examples

$y = f (x + 5)$

Step 1

Simplify $f (x + 5)$ .

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Step 1.1

Apply the distributive property.

$y = f x + f \cdot 5$

Step 1.2

Move $5$ to the left of $f$ .

$y = f x + 5 f$

Step 2

Choose any values for $x$ and $y$ that are in the domain to plug into the equation.

Step 3

Choose $0$ to substitute in for $x$ and $1$ to substitute for $z$ .

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Step 3.1

Remove parentheses.

$y = f (0) + 5 f$

Step 3.2

Simplify $f (0) + 5 f$ .

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Step 3.2.1

Multiply $f$ by $0$ .

$y = 0 + 5 f$

Step 3.2.2

Add $0$ and $5 f$ .

$y = 5 f$

Step 3.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(0, 5 f, 1)$

Step 4

Choose $1$ to substitute in for $x$ and $2$ to substitute for $z$ .

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Step 4.1

Remove parentheses.

$y = f (1) + 5 f$

Step 4.2

Simplify $f (1) + 5 f$ .

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Step 4.2.1

Multiply $f$ by $1$ .

$y = f + 5 f$

Step 4.2.2

Add $f$ and $5 f$ .

$y = 6 f$

Step 4.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(1, 6 f, 2)$

Step 5

Choose $2$ to substitute in for $x$ and $3$ to substitute for $z$ .

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Step 5.1

Remove parentheses.

$y = f (2) + 5 f$

Step 5.2

Simplify $f (2) + 5 f$ .

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Step 5.2.1

Move $2$ to the left of $f$ .

$y = 2 f + 5 f$

Step 5.2.2

Add $2 f$ and $5 f$ .

$y = 7 f$

Step 5.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(2, 7 f, 3)$

Step 6

These are three possible solutions to the equation.

$(0, 5 f, 1), (1, 6 f, 2), (2, 7 f, 3)$