Trigonometry Examples

$\log_{2} (2 x^{2} + 24 x + 72)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$2 x^{2} + 24 x + 72 = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Factor the left side of the equation.

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Step 1.2.1.1

Factor $2$ out of $2 x^{2} + 24 x + 72$ .

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Step 1.2.1.1.1

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 24 x + 72 = 0$

Step 1.2.1.1.2

Factor $2$ out of $24 x$ .

$2 (x^{2}) + 2 (12 x) + 72 = 0$

Step 1.2.1.1.3

Factor $2$ out of $72$ .

$2 x^{2} + 2 (12 x) + 2 \cdot 36 = 0$

Step 1.2.1.1.4

Factor $2$ out of $2 x^{2} + 2 (12 x)$ .

$2 (x^{2} + 12 x) + 2 \cdot 36 = 0$

Step 1.2.1.1.5

Factor $2$ out of $2 (x^{2} + 12 x) + 2 \cdot 36$ .

$2 (x^{2} + 12 x + 36) = 0$

Step 1.2.1.2

Factor using the perfect square rule.

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Step 1.2.1.2.1

Rewrite $36$ as $6^{2}$ .

$2 (x^{2} + 12 x + 6^{2}) = 0$

Step 1.2.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Step 1.2.1.2.3

Rewrite the polynomial.

$2 (x^{2} + 2 \cdot x \cdot 6 + 6^{2}) = 0$

Step 1.2.1.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 6$ .

$2 {(x + 6)}^{2} = 0$

Step 1.2.2

Divide each term in $2 {(x + 6)}^{2} = 0$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 {(x + 6)}^{2} = 0$ by $2$ .

$\frac{2 {(x + 6)}^{2}}{2} = \frac{0}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{2 {(x + 6)}^{2}}{2} = \frac{0}{2}$

Step 1.2.2.2.1.2

Divide ${(x + 6)}^{2}$ by $1$ .

${(x + 6)}^{2} = \frac{0}{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $0$ by $2$ .

${(x + 6)}^{2} = 0$

Step 1.2.3

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 1.2.4

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 1.3

The vertical asymptote occurs at $x = - 6$ .

Vertical Asymptote: $x = - 6$

Step 2

Find the point at $x = - 5$ .

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Step 2.1

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = \log_{2} (2 {(- 5)}^{2} + 24 (- 5) + 72)$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

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

$f (- 5) = \log_{2} (2 \cdot 25 + 24 (- 5) + 72)$

Step 2.2.1.2

Multiply $2$ by $25$ .

$f (- 5) = \log_{2} (50 + 24 (- 5) + 72)$

Step 2.2.1.3

Multiply $24$ by $- 5$ .

$f (- 5) = \log_{2} (50 - 120 + 72)$

Step 2.2.2

Simplify by adding and subtracting.

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Step 2.2.2.1

Subtract $120$ from $50$ .

$f (- 5) = \log_{2} (- 70 + 72)$

Step 2.2.2.2

Add $- 70$ and $72$ .

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

Step 2.2.3

Logarithm base $2$ of $2$ is $1$ .

$f (- 5) = 1$

Step 2.2.4

The final answer is $1$ .

$1$

Step 2.3

Convert $1$ to decimal.

$y = 1$

Step 3

Find the point at $x = - 4$ .

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Step 3.1

Replace the variable $x$ with $- 4$ in the expression.

$f (- 4) = \log_{2} (2 {(- 4)}^{2} + 24 (- 4) + 72)$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Raise $- 4$ to the power of $2$ .

$f (- 4) = \log_{2} (2 \cdot 16 + 24 (- 4) + 72)$

Step 3.2.1.2

Multiply $2$ by $16$ .

$f (- 4) = \log_{2} (32 + 24 (- 4) + 72)$

Step 3.2.1.3

Multiply $24$ by $- 4$ .

$f (- 4) = \log_{2} (32 - 96 + 72)$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Subtract $96$ from $32$ .

$f (- 4) = \log_{2} (- 64 + 72)$

Step 3.2.2.2

Add $- 64$ and $72$ .

$f (- 4) = \log_{2} (8)$

Step 3.2.3

Logarithm base $2$ of $8$ is $3$ .

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Step 3.2.3.1

Rewrite as an equation.

$\log_{2} (8) = x$

Step 3.2.3.2

Rewrite $\log_{2} (8) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$2^{x} = 8$

Step 3.2.3.3

Create equivalent expressions in the equation that all have equal bases.

$2^{x} = 2^{3}$

Step 3.2.3.4

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

$x = 3$

Step 3.2.3.5

The variable $x$ is equal to $3$ .

$f (- 4) = 3$

Step 3.2.4

The final answer is $3$ .

$3$

Step 3.3

Convert $3$ to decimal.

$y = 3$

Step 4

Find the point at $x = - 2$ .

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Step 4.1

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

$f (- 2) = \log_{2} (2 {(- 2)}^{2} + 24 (- 2) + 72)$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

$f (- 2) = \log_{2} (2 \cdot 4 + 24 (- 2) + 72)$

Step 4.2.1.2

Multiply $2$ by $4$ .

$f (- 2) = \log_{2} (8 + 24 (- 2) + 72)$

Step 4.2.1.3

Multiply $24$ by $- 2$ .

$f (- 2) = \log_{2} (8 - 48 + 72)$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Subtract $48$ from $8$ .

$f (- 2) = \log_{2} (- 40 + 72)$

Step 4.2.2.2

Add $- 40$ and $72$ .

$f (- 2) = \log_{2} (32)$

Step 4.2.3

Logarithm base $2$ of $32$ is $5$ .

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Step 4.2.3.1

Rewrite as an equation.

$\log_{2} (32) = x$

Step 4.2.3.2

Rewrite $\log_{2} (32) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$2^{x} = 32$

Step 4.2.3.3

Create equivalent expressions in the equation that all have equal bases.

$2^{x} = 2^{5}$

Step 4.2.3.4

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

$x = 5$

Step 4.2.3.5

The variable $x$ is equal to $5$ .

$f (- 2) = 5$

Step 4.2.4

The final answer is $5$ .

$5$

Step 4.3

Convert $5$ to decimal.

$y = 5$

Step 5

The log function can be graphed using the vertical asymptote at $x = - 6$ and the points $(- 5, 1), (- 4, 3), (- 2, 5)$ .

Vertical Asymptote: $x = - 6$

$\begin{array}{c} x & y \\ - 5 & 1 \\ - 4 & 3 \\ - 2 & 5 \end{array}$

Step 6