QuestionAnalyze the function $R(x)=\frac{x^{2}+x-30}{x-6}$ . Find its domain, vertical asymptote, and horizontal/oblique asymptote.

Studdy Solution

STEP 1

Assumptions1. The function is given as $R(x)=\frac{x^{}+x-30}{x-6}$ . We are asked to find the domain of $R(x)$ , the equation of the vertical asymptote(s), and the equation of the horizontal or oblique asymptote.

STEP 2

First, let's find the domain of $R(x)$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a rational function, the domain is all real numbers except for those which make the denominator zero.

STEP 3

To find the values that make the denominator zero, we set the denominator equal to zero and solve for x.
$x -6 =0$

STEP 4

olving the equation gives us the x-value that is not in the domain of $R(x)$ .
$x =6$

STEP 5

So, the domain of $R(x)$ is all real numbers except. In set notation, this is written as $\{x \mid x \neq\}$ .

STEP 6

Next, let's find the vertical asymptote(s) of $R(x)$ . A vertical asymptote of a function is a vertical line $x = a$ where the function approaches infinity (or negative infinity) as $x$ approaches $a$ .

STEP 7

For a rational function, the vertical asymptotes are the values of x that make the denominator zero (and are not also zeros of the numerator). We have already found that $x =6$ makes the denominator zero, so $x =6$ is the vertical asymptote.

STEP 8

Finally, let's find the horizontal or oblique asymptote of $R(x)$ . A horizontal asymptote of a function is a horizontal line $y = b$ where the function approaches $b$ as $x$ approaches infinity (or negative infinity). An oblique asymptote is a slanted line that the graph of the function approaches as $x$ approaches infinity (or negative infinity).

STEP 9

For a rational function, if the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is one more than the degree of the denominator, there is an oblique asymptote, which can be found by performing polynomial division.

STEP 10

In this case, the degree of the numerator is2 and the degree of the denominator is, so there is an oblique asymptote. We can find it by performing polynomial division.

STEP 11

Performing the division $x^{}+x-30$ divided by $x-6$ gives us $x+7$ with a remainder of $-$ .

STEP 12

The oblique asymptote is then the quotient $x+7$ .
So, the answers are(a) The domain of $R(x)$ is $\{x \mid x \neq6\}$ . (b) The equation of the vertical asymptote is $x=6$ . (c) The equation of the oblique asymptote is $y=x+7$ .

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QuestionAnalyze the function $R(x)=\frac{x^{2}+x-30}{x-6}$ . Find its domain, vertical asymptote, and horizontal/oblique asymptote.

Studdy Solution

STEP 1

Assumptions1. The function is given as $R(x)=\frac{x^{}+x-30}{x-6}$ . We are asked to find the domain of $R(x)$ , the equation of the vertical asymptote(s), and the equation of the horizontal or oblique asymptote.

STEP 2

First, let's find the domain of $R(x)$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a rational function, the domain is all real numbers except for those which make the denominator zero.

STEP 3

To find the values that make the denominator zero, we set the denominator equal to zero and solve for x.
$x -6 =0$

STEP 4

olving the equation gives us the x-value that is not in the domain of $R(x)$ .
$x =6$

STEP 5

So, the domain of $R(x)$ is all real numbers except. In set notation, this is written as $\{x \mid x \neq\}$ .

STEP 6

Next, let's find the vertical asymptote(s) of $R(x)$ . A vertical asymptote of a function is a vertical line $x = a$ where the function approaches infinity (or negative infinity) as $x$ approaches $a$ .

STEP 7

For a rational function, the vertical asymptotes are the values of x that make the denominator zero (and are not also zeros of the numerator). We have already found that $x =6$ makes the denominator zero, so $x =6$ is the vertical asymptote.

STEP 8

STEP 9

STEP 10

In this case, the degree of the numerator is2 and the degree of the denominator is, so there is an oblique asymptote. We can find it by performing polynomial division.

STEP 11

Performing the division $x^{}+x-30$ divided by $x-6$ gives us $x+7$ with a remainder of $-$ .

STEP 12

The oblique asymptote is then the quotient $x+7$ .
So, the answers are(a) The domain of $R(x)$ is $\{x \mid x \neq6\}$ . (b) The equation of the vertical asymptote is $x=6$ . (c) The equation of the oblique asymptote is $y=x+7$ .

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QuestionAnalyze the function R(x)=x2+x−30x−6R(x)=\frac{x^{2}+x-30}{x-6}R(x)=x−6x2+x−30​. Find its domain, vertical asymptote, and horizontal/oblique asymptote.

Studdy Solution

STEP 1

Assumptions1. The function is given as R(x)=x+x−30x−6R(x)=\frac{x^{}+x-30}{x-6}R(x)=x−6x+x−30​ . We are asked to find the domain of R(x)R(x)R(x), the equation of the vertical asymptote(s), and the equation of the horizontal or oblique asymptote.

STEP 2

First, let's find the domain of R(x)R(x)R(x). The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a rational function, the domain is all real numbers except for those which make the denominator zero.

STEP 3

To find the values that make the denominator zero, we set the denominator equal to zero and solve for x.x−6=0x -6 =0x−6=0

STEP 4

olving the equation gives us the x-value that is not in the domain of R(x)R(x)R(x).x=6x =6x=6

STEP 5

So, the domain of R(x)R(x)R(x) is all real numbers except. In set notation, this is written as {x∣x≠}\{x \mid x \neq\}{x∣x=}.

STEP 6

Next, let's find the vertical asymptote(s) of R(x)R(x)R(x). A vertical asymptote of a function is a vertical line x=ax = ax=a where the function approaches infinity (or negative infinity) as xxx approaches aaa.

STEP 7

For a rational function, the vertical asymptotes are the values of x that make the denominator zero (and are not also zeros of the numerator). We have already found that x=6x =6x=6 makes the denominator zero, so x=6x =6x=6 is the vertical asymptote.

STEP 8

STEP 9

STEP 10

In this case, the degree of the numerator is2 and the degree of the denominator is, so there is an oblique asymptote. We can find it by performing polynomial division.

STEP 11

Performing the division x+x−30x^{}+x-30x+x−30 divided by x−6x-6x−6 gives us x+7x+7x+7 with a remainder of −-−.

STEP 12

The oblique asymptote is then the quotient x+7x+7x+7.So, the answers are(a) The domain of R(x)R(x)R(x) is {x∣x≠6}\{x \mid x \neq6\}{x∣x=6}. (b) The equation of the vertical asymptote is x=6x=6x=6. (c) The equation of the oblique asymptote is y=x+7y=x+7y=x+7.

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QuestionAnalyze the function $R(x)=\frac{x^{2}+x-30}{x-6}$ . Find its domain, vertical asymptote, and horizontal/oblique asymptote.

Assumptions1. The function is given as $R(x)=\frac{x^{}+x-30}{x-6}$ . We are asked to find the domain of $R(x)$ , the equation of the vertical asymptote(s), and the equation of the horizontal or oblique asymptote.

First, let's find the domain of $R(x)$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output. For a rational function, the domain is all real numbers except for those which make the denominator zero.

To find the values that make the denominator zero, we set the denominator equal to zero and solve for x.
$x -6 =0$

olving the equation gives us the x-value that is not in the domain of $R(x)$ .
$x =6$

So, the domain of $R(x)$ is all real numbers except. In set notation, this is written as $\{x \mid x \neq\}$ .

Next, let's find the vertical asymptote(s) of $R(x)$ . A vertical asymptote of a function is a vertical line $x = a$ where the function approaches infinity (or negative infinity) as $x$ approaches $a$ .

For a rational function, the vertical asymptotes are the values of x that make the denominator zero (and are not also zeros of the numerator). We have already found that $x =6$ makes the denominator zero, so $x =6$ is the vertical asymptote.

Performing the division $x^{}+x-30$ divided by $x-6$ gives us $x+7$ with a remainder of $-$ .

The oblique asymptote is then the quotient $x+7$ .
So, the answers are(a) The domain of $R(x)$ is $\{x \mid x \neq6\}$ . (b) The equation of the vertical asymptote is $x=6$ . (c) The equation of the oblique asymptote is $y=x+7$ .