Graph f(x)=2 log base 5 of x+3-1

$f (x) = 2 \log_{5} (x + 3) - 1$

Find the asymptotes.

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Find where the expression $\log_{5} ({(x + 3)}^{2}) - 1$ is undefined.

$x = - 3$

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).

There are no horizontal asymptotes because $Q (x)$ is $1$ .

No Horizontal Asymptotes

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes: $x = - 3$

No Horizontal Asymptotes

Vertical Asymptotes: $x = - 3$

No Horizontal Asymptotes

Find the point at $x = - 2$ .

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

$f (- 2) = 2 \log_{5} ((- 2) + 3) - 1$

Simplify the result.

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Simplify each term.

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Add $- 2$ and $3$ .

$f (- 2) = 2 \log_{5} (1) - 1$

Logarithm base $5$ of $1$ is $0$ .

$f (- 2) = 2 \cdot 0 - 1$

Multiply $2$ by $0$ .

$f (- 2) = 0 - 1$

Subtract $1$ from $0$ .

$f (- 2) = - 1$

The final answer is $- 1$ .

$- 1$

Convert $- 1$ to decimal.

$y = - 1$

Find the point at $x = 2$ .

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

$f (2) = 2 \log_{5} ((2) + 3) - 1$

Simplify the result.

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Simplify each term.

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Add $2$ and $3$ .

$f (2) = 2 \log_{5} (5) - 1$

Logarithm base $5$ of $5$ is $1$ .

$f (2) = 2 \cdot 1 - 1$

Multiply $2$ by $1$ .

$f (2) = 2 - 1$

Subtract $1$ from $2$ .

$f (2) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = - 1$ .

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

$f (- 1) = 2 \log_{5} ((- 1) + 3) - 1$

Simplify the result.

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Simplify each term.

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Add $- 1$ and $3$ .

$f (- 1) = 2 \log_{5} (2) - 1$

Simplify $2 \log_{5} (2)$ by moving $2$ inside the logarithm.

$f (- 1) = \log_{5} (2^{2}) - 1$

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

$f (- 1) = \log_{5} (4) - 1$

The final answer is $\log_{5} (4) - 1$ .

$\log_{5} (4) - 1$

Convert $\log_{5} (4) - 1$ to decimal.

$y = - 0.13864688$

The log function can be graphed using the vertical asymptote at $x = - 3$ and the points $(- 2, - 1), (2, 1), (- 1, - 0.13864688)$ .

Vertical Asymptote: $x = - 3$

$\begin{array}{c} x & y \\ - 2 & - 1 \\ - 1 & - 0.139 \\ 2 & 1 \end{array}$

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