Finite Math Examples

$h (x) = {(x - 1)}^{2} + 2$

Step 1

A rational function is any function which can be written as the ratio of two polynomial functions where the denominator is not $0$ .

$h (x) = {(x - 1)}^{2} + 2$ is a rational function

Step 2

$h (x) = {(x - 1)}^{2} + 2$ can be written as $h (x) = \frac{{(x - 1)}^{2} + 2}{1}$ .

Step 3

A rational function is proper when the degree of the numerator is less than the degree of the denominator, otherwise it is improper.

Degree of numerator is less than the degree of denominator implies a proper function

Degree of numerator is greater than the degree of denominator implies an improper function

Degree of numerator is equal to the degree of denominator implies an improper function

Step 4

Find the degree of the numerator.

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Step 4.1

Simplify and reorder the polynomial.

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Step 4.1.1

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$(x - 1) (x - 1)$

Step 4.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 4.1.2.1

Apply the distributive property.

$x (x - 1) - 1 (x - 1)$

Step 4.1.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 1 (x - 1)$

Step 4.1.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1$

Step 4.1.3

Simplify and combine like terms.

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Step 4.1.3.1

Simplify each term.

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Step 4.1.3.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1$

Step 4.1.3.1.2

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1$

Step 4.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - 1 x - 1 \cdot - 1$

Step 4.1.3.1.4

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - x - 1 \cdot - 1$

Step 4.1.3.1.5

Multiply $- 1$ by $- 1$ .

$x^{2} - x - x + 1$

Step 4.1.3.2

Subtract $x$ from $- x$ .

$x^{2} - 2 x + 1$

Step 4.2

The largest exponent is the degree of the polynomial.

$2$

Step 5

The expression is constant, which means it can be rewritten with a factor of $x^{0}$ . The degree is the largest exponent on the variable.

$0$

Step 6

The degree of the numerator $2$ is greater than the degree of the denominator $0$ .

$2 > 0$

Step 7

The degree of the numerator is greater than the degree of the denominator, which means that $h (x)$ is an improper function.

Improper