Precalculus Examples

$f (x) = \frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$

Step 1

Find where the expression $\frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ is undefined.

$x = \ln (2), x = \ln (3)$

Step 2

Since $\frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ $\to$ $\infty$ as $x$ $\to$ $\ln (2)$ from the left and $\frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ $\to$ $- \infty$ as $x$ $\to$ $\ln (2)$ from the right, then $x = \ln (2)$ is a vertical asymptote.

$x = \ln (2)$

Step 3

Since $\frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ $\to$ $- \infty$ as $x$ $\to$ $\ln (3)$ from the left and $\frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ $\to$ $\infty$ as $x$ $\to$ $\ln (3)$ from the right, then $x = \ln (3)$ is a vertical asymptote.

$x = \ln (3)$

Step 4

List all of the vertical asymptotes:

$x = \ln (2), \ln (3)$

Step 5

Evaluate $lim_{x \to - \infty} \frac{3 e^{2 x} + 2}{e^{2 x} - 5 e^{x} + 6}$ to find the horizontal asymptote.

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Step 5.1

Evaluate the limit.

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Step 5.1.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- \infty$ .

$\frac{lim_{x \to - \infty} 3 e^{2 x} + 2}{lim_{x \to - \infty} e^{2 x} - 5 e^{x} + 6}$

Step 5.1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- \infty$ .

$\frac{lim_{x \to - \infty} 3 e^{2 x} + lim_{x \to - \infty} 2}{lim_{x \to - \infty} e^{2 x} - 5 e^{x} + 6}$

Step 5.1.3

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{3 lim_{x \to - \infty} e^{2 x} + lim_{x \to - \infty} 2}{lim_{x \to - \infty} e^{2 x} - 5 e^{x} + 6}$

Step 5.2

Since the exponent $2 x$ approaches $- \infty$ , the quantity $e^{2 x}$ approaches $0$ .

$\frac{3 \cdot 0 + lim_{x \to - \infty} 2}{lim_{x \to - \infty} e^{2 x} - 5 e^{x} + 6}$

Step 5.3

Evaluate the limit.

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Step 5.3.1

Evaluate the limit of $2$ which is constant as $x$ approaches $- \infty$ .

$\frac{3 \cdot 0 + 2}{lim_{x \to - \infty} e^{2 x} - 5 e^{x} + 6}$

Step 5.3.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- \infty$ .

$\frac{3 \cdot 0 + 2}{lim_{x \to - \infty} e^{2 x} - lim_{x \to - \infty} 5 e^{x} + lim_{x \to - \infty} 6}$

Step 5.4

Since the exponent $2 x$ approaches $- \infty$ , the quantity $e^{2 x}$ approaches $0$ .

$\frac{3 \cdot 0 + 2}{0 - lim_{x \to - \infty} 5 e^{x} + lim_{x \to - \infty} 6}$

Step 5.5

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{3 \cdot 0 + 2}{0 - 5 lim_{x \to - \infty} e^{x} + lim_{x \to - \infty} 6}$

Step 5.6

Since the exponent $x$ approaches $- \infty$ , the quantity $e^{x}$ approaches $0$ .

$\frac{3 \cdot 0 + 2}{0 - 5 \cdot 0 + lim_{x \to - \infty} 6}$

Step 5.7

Evaluate the limit.

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Step 5.7.1

Evaluate the limit of $6$ which is constant as $x$ approaches $- \infty$ .

$\frac{3 \cdot 0 + 2}{0 - 5 \cdot 0 + 6}$

Step 5.7.2

Simplify the answer.

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Step 5.7.2.1

Cancel the common factor of $3 \cdot 0 + 2$ and $0 - 5 \cdot 0 + 6$ .

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Step 5.7.2.1.1

Rewrite $0$ as $- 1 (0)$ .

$\frac{3 \cdot 0 + 2}{- 1 (0) - 5 \cdot 0 + 6}$

Step 5.7.2.1.2

Factor $- 1$ out of $- 5 \cdot 0$ .

$\frac{3 \cdot 0 + 2}{- 1 (0) - (5 \cdot 0) + 6}$

Step 5.7.2.1.3

Factor $- 1$ out of $- 1 (0) - (5 \cdot 0)$ .

$\frac{3 \cdot 0 + 2}{- 1 (0 + 5 \cdot 0) + 6}$

Step 5.7.2.1.4

Rewrite $6$ as $- 1 (- 6)$ .

$\frac{3 \cdot 0 + 2}{- 1 (0 + 5 \cdot 0) - 1 \cdot - 6}$

Step 5.7.2.1.5

Factor $- 1$ out of $- 1 (0 + 5 \cdot 0) - 1 \cdot - 6$ .

$\frac{3 \cdot 0 + 2}{- 1 (0 + 5 \cdot 0 - 6)}$

Step 5.7.2.1.6

Reorder terms.

$\frac{3 \cdot 0 + 2}{- 1 (0 + 0 \cdot 5 - 6)}$

Step 5.7.2.1.7

Factor $2$ out of $3 \cdot 0$ .

$\frac{2 (3 \cdot 0) + 2}{- 1 (0 + 0 \cdot 5 - 6)}$

Step 5.7.2.1.8

Factor $2$ out of $2$ .

$\frac{2 (3 \cdot 0) + 2 (1)}{- 1 (0 + 0 \cdot 5 - 6)}$

Step 5.7.2.1.9

Factor $2$ out of $2 (3 \cdot 0) + 2 (1)$ .

$\frac{2 (3 \cdot 0 + 1)}{- 1 (0 + 0 \cdot 5 - 6)}$

Step 5.7.2.1.10

Cancel the common factors.

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Step 5.7.2.1.10.1

Factor $2$ out of $- 1 (0 + 0 \cdot 5 - 6)$ .

$\frac{2 (3 \cdot 0 + 1)}{2 (- 1 (0 + 0 \cdot 5 - 3))}$

Step 5.7.2.1.10.2

Cancel the common factor.

$\frac{2 (3 \cdot 0 + 1)}{2 (- 1 (0 + 0 \cdot 5 - 3))}$

Step 5.7.2.1.10.3

Rewrite the expression.

$\frac{3 \cdot 0 + 1}{- 1 (0 + 0 \cdot 5 - 3)}$

Step 5.7.2.2

Simplify the numerator.

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Step 5.7.2.2.1

Multiply $3$ by $0$ .

$\frac{0 + 1}{- 1 (0 + 0 \cdot 5 - 3)}$

Step 5.7.2.2.2

Add $0$ and $1$ .

$\frac{1}{- 1 (0 + 0 \cdot 5 - 3)}$

Step 5.7.2.3

Simplify the denominator.

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Step 5.7.2.3.1

Multiply $0$ by $5$ .

$\frac{1}{- 1 (0 + 0 - 3)}$

Step 5.7.2.3.2

Add $0$ and $0$ .

$\frac{1}{- 1 (0 - 3)}$

Step 5.7.2.3.3

Subtract $3$ from $0$ .

$\frac{1}{- 1 \cdot - 3}$

Step 5.7.2.4

Multiply $- 1$ by $- 3$ .

$\frac{1}{3}$

Step 6

List the horizontal asymptotes:

$y = \frac{1}{3}$

Step 7

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

Step 8

This is the set of all asymptotes.

Vertical Asymptotes: $x = \ln (2), \ln (3)$

Horizontal Asymptotes: $y = \frac{1}{3}$

No Oblique Asymptotes

Step 9