Finite Math Examples

$y = \frac{5 \ln (x + 5)}{x^{2}}$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{\ln ({(x + 5)}^{5})}{x^{2}}$ is undefined.

$x \leq - 5, x = 0$

Step 1.2

Since $\frac{\ln ({(x + 5)}^{5})}{x^{2}}$ $\to$ $\infty$ as $x$ $\to$ $- 5$ from the left and $\frac{\ln ({(x + 5)}^{5})}{x^{2}}$ $\to$ $- \infty$ as $x$ $\to$ $- 5$ from the right, then $x = - 5$ is a vertical asymptote.

$x = - 5$

Step 1.3

Since $\frac{\ln ({(x + 5)}^{5})}{x^{2}}$ $\to$ $\infty$ as $x$ $\to$ $0$ from the left and $\frac{\ln ({(x + 5)}^{5})}{x^{2}}$ $\to$ $\infty$ as $x$ $\to$ $0$ from the right, then $x = 0$ is a vertical asymptote.

$x = 0$

Step 1.4

List all of the vertical asymptotes:

$x = - 5, 0$

Step 1.5

Ignoring the logarithm, consider the rational function $R (x) = \frac{a x^{n}}{b x^{m}}$ where $n$ is the degree of the numerator and $m$ is the degree of the denominator.

1. If $n < m$ , then the x-axis, $y = 0$ , is the horizontal asymptote.

2. If $n = m$ , then the horizontal asymptote is the line $y = \frac{a}{b}$ .

3. If $n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

Step 1.6

Find $n$ and $m$ .

$n = 0$

$m = 2$

Step 1.7

Since $n < m$ , the x-axis, $y = 0$ , is the horizontal asymptote.

$y = 0$

Step 1.8

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.9

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 5, 0$

Horizontal Asymptotes: $y = 0$

Vertical Asymptotes: $x = - 5, 0$

Horizontal Asymptotes: $y = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

$f (1) = \frac{5 \ln ((1) + 5)}{{(1)}^{2}}$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify $5 \ln (1 + 5)$ by moving $5$ inside the logarithm.

$f (1) = \frac{\ln ({(1 + 5)}^{5})}{1^{2}}$

Step 2.2.2

One to any power is one.

$f (1) = \frac{\ln ({(1 + 5)}^{5})}{1}$

Step 2.2.3

Simplify the numerator.

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Step 2.2.3.1

Add $1$ and $5$ .

$f (1) = \frac{\ln (6^{5})}{1}$

Step 2.2.3.2

Raise $6$ to the power of $5$ .

$f (1) = \frac{\ln (7776)}{1}$

Step 2.2.4

Divide $\ln (7776)$ by $1$ .

$f (1) = \ln (7776)$

Step 2.2.5

The final answer is $\ln (7776)$ .

$\ln (7776)$

Step 2.3

Convert $\ln (7776)$ to decimal.

$y = 8.95879734$

Step 3

Find the point at $x = 2$ .

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Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \frac{5 \ln ((2) + 5)}{{(2)}^{2}}$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify $5 \ln (2 + 5)$ by moving $5$ inside the logarithm.

$f (2) = \frac{\ln ({(2 + 5)}^{5})}{2^{2}}$

Step 3.2.2

Raise $2$ to the power of $2$ .

$f (2) = \frac{\ln ({(2 + 5)}^{5})}{4}$

Step 3.2.3

Simplify the numerator.

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Step 3.2.3.1

Add $2$ and $5$ .

$f (2) = \frac{\ln (7^{5})}{4}$

Step 3.2.3.2

Raise $7$ to the power of $5$ .

$f (2) = \frac{\ln (16807)}{4}$

Step 3.2.4

Rewrite $\frac{\ln (16807)}{4}$ as $\frac{1}{4} \ln (16807)$ .

$f (2) = \frac{1}{4} \cdot \ln (16807)$

Step 3.2.5

Simplify $\frac{1}{4} \ln (16807)$ by moving $\frac{1}{4}$ inside the logarithm.

$f (2) = \ln (16807^{\frac{1}{4}})$

Step 3.2.6

The final answer is $\ln (16807^{\frac{1}{4}})$ .

$\ln (16807^{\frac{1}{4}})$

Step 3.3

Convert $\ln (16807^{\frac{1}{4}})$ to decimal.

$y = 2.43238768$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = \frac{5 \ln ((3) + 5)}{{(3)}^{2}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify $5 \ln (3 + 5)$ by moving $5$ inside the logarithm.

$f (3) = \frac{\ln ({(3 + 5)}^{5})}{3^{2}}$

Step 4.2.2

Raise $3$ to the power of $2$ .

$f (3) = \frac{\ln ({(3 + 5)}^{5})}{9}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Add $3$ and $5$ .

$f (3) = \frac{\ln (8^{5})}{9}$

Step 4.2.3.2

Raise $8$ to the power of $5$ .

$f (3) = \frac{\ln (32768)}{9}$

Step 4.2.4

Rewrite $\ln (32768)$ as $\ln (2^{15})$ .

$f (3) = \frac{\ln (2^{15})}{9}$

Step 4.2.5

Expand $\ln (2^{15})$ by moving $15$ outside the logarithm.

$f (3) = \frac{15 \ln (2)}{9}$

Step 4.2.6

Cancel the common factor of $15$ and $9$ .

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Step 4.2.6.1

Factor $3$ out of $15 \ln (2)$ .

$f (3) = \frac{3 (5 \ln (2))}{9}$

Step 4.2.6.2

Cancel the common factors.

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Step 4.2.6.2.1

Factor $3$ out of $9$ .

$f (3) = \frac{3 (5 \ln (2))}{3 (3)}$

Step 4.2.6.2.2

Cancel the common factor.

$f (3) = \frac{3 (5 \ln (2))}{3 \cdot 3}$

Step 4.2.6.2.3

Rewrite the expression.

$f (3) = \frac{5 \ln (2)}{3}$

Step 4.2.7

Simplify $5 \ln (2)$ by moving $5$ inside the logarithm.

$f (3) = \frac{\ln (2^{5})}{3}$

Step 4.2.8

Raise $2$ to the power of $5$ .

$f (3) = \frac{\ln (32)}{3}$

Step 4.2.9

Rewrite $\frac{\ln (32)}{3}$ as $\frac{1}{3} \ln (32)$ .

$f (3) = \frac{1}{3} \cdot \ln (32)$

Step 4.2.10

Simplify $\frac{1}{3} \ln (32)$ by moving $\frac{1}{3}$ inside the logarithm.

$f (3) = \ln (32^{\frac{1}{3}})$

Step 4.2.11

The final answer is $\ln (32^{\frac{1}{3}})$ .

$\ln (32^{\frac{1}{3}})$

Step 4.3

Convert $\ln (32^{\frac{1}{3}})$ to decimal.

$y = 1.1552453$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 5, 0$ and the points $(1, 8.95879734), (2, 2.43238768), (3, 1.1552453)$ .

Vertical Asymptote: $x = - 5, 0$

$\begin{array}{c} x & y \\ 1 & 8.959 \\ 2 & 2.432 \\ 3 & 1.155 \end{array}$

Step 6