Calculus Examples

f (x) = ln (x^{2} - 2 x + 5)

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

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$x^{2} - 2 x + 5 = 0$

Step 1.2

Solve for

$x$ .

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Step 1.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.2.2

Substitute the values

$a = 1$ ,

$b = - 2$ , and

$c = 5$ into the quadratic formula and solve for

$x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 1.2.3

Simplify.

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Step 1.2.3.1

Simplify the numerator.

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Step 1.2.3.1.1

Raise

$- 2$ to the power of

$2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 1.2.3.1.2

Multiply

$- 4 \cdot 1 \cdot 5$ .

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Step 1.2.3.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Step 1.2.3.1.2.2

Multiply

$- 4$ by

$5$ .

$x = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 1.2.3.1.3

Subtract

$20$ from

$4$ .

$x = \frac{2 \pm \sqrt{- 16}}{2 \cdot 1}$

Step 1.2.3.1.4

Rewrite

$- 16$ as

$- 1 (16)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Step 1.2.3.1.5

Rewrite

$\sqrt{- 1 (16)}$ as

$\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.3.1.6

Rewrite

$\sqrt{- 1}$ as

$i$ .

$x = \frac{2 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.3.1.7

Rewrite

$16$ as

$4^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Step 1.2.3.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 4}{2 \cdot 1}$

Step 1.2.3.1.9

Move

$4$ to the left of

$i$ .

$x = \frac{2 \pm 4 i}{2 \cdot 1}$

Step 1.2.3.2

Multiply

$2$ by

$1$ .

$x = \frac{2 \pm 4 i}{2}$

Step 1.2.3.3

Simplify

$\frac{2 \pm 4 i}{2}$ .

$x = 1 \pm 2 i$

Step 1.2.4

Simplify the expression to solve for the

$+$ portion of the

$\pm$ .

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Step 1.2.4.1

Simplify the numerator.

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Step 1.2.4.1.1

Raise

$- 2$ to the power of

$2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 1.2.4.1.2

Multiply

$- 4 \cdot 1 \cdot 5$ .

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Step 1.2.4.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Step 1.2.4.1.2.2

Multiply

$- 4$ by

$5$ .

$x = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 1.2.4.1.3

Subtract

$20$ from

$4$ .

$x = \frac{2 \pm \sqrt{- 16}}{2 \cdot 1}$

Step 1.2.4.1.4

Rewrite

$- 16$ as

$- 1 (16)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Step 1.2.4.1.5

Rewrite

$\sqrt{- 1 (16)}$ as

$\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.4.1.6

Rewrite

$\sqrt{- 1}$ as

$i$ .

$x = \frac{2 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.4.1.7

Rewrite

$16$ as

$4^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Step 1.2.4.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 4}{2 \cdot 1}$

Step 1.2.4.1.9

Move

$4$ to the left of

$i$ .

$x = \frac{2 \pm 4 i}{2 \cdot 1}$

Step 1.2.4.2

Multiply

$2$ by

$1$ .

$x = \frac{2 \pm 4 i}{2}$

Step 1.2.4.3

Simplify

$\frac{2 \pm 4 i}{2}$ .

$x = 1 \pm 2 i$

Step 1.2.4.4

Change the

$\pm$ to

$+$ .

$x = 1 + 2 i$

Step 1.2.5

Simplify the expression to solve for the

$-$ portion of the

$\pm$ .

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Step 1.2.5.1

Simplify the numerator.

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Step 1.2.5.1.1

Raise

$- 2$ to the power of

$2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 1.2.5.1.2

Multiply

$- 4 \cdot 1 \cdot 5$ .

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Step 1.2.5.1.2.1

Multiply

$- 4$ by

$1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Step 1.2.5.1.2.2

Multiply

$- 4$ by

$5$ .

$x = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 1.2.5.1.3

Subtract

$20$ from

$4$ .

$x = \frac{2 \pm \sqrt{- 16}}{2 \cdot 1}$

Step 1.2.5.1.4

Rewrite

$- 16$ as

$- 1 (16)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Step 1.2.5.1.5

Rewrite

$\sqrt{- 1 (16)}$ as

$\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.5.1.6

Rewrite

$\sqrt{- 1}$ as

$i$ .

$x = \frac{2 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Step 1.2.5.1.7

Rewrite

$16$ as

$4^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Step 1.2.5.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 4}{2 \cdot 1}$

Step 1.2.5.1.9

Move

$4$ to the left of

$i$ .

$x = \frac{2 \pm 4 i}{2 \cdot 1}$

Step 1.2.5.2

Multiply

$2$ by

$1$ .

$x = \frac{2 \pm 4 i}{2}$

Step 1.2.5.3

Simplify

$\frac{2 \pm 4 i}{2}$ .

$x = 1 \pm 2 i$

Step 1.2.5.4

Change the

$\pm$ to

$-$ .

$x = 1 - 2 i$

Step 1.2.6

The final answer is the combination of both solutions.

$x = 1 + 2 i, 1 - 2 i$

Step 1.3

The vertical asymptote occurs at

$x = 1 + 2 i, x = 1 - 2 i$ .

Vertical Asymptote:

$x = 1 + 2 i, x = 1 - 2 i$

Vertical Asymptote:

$x = 1 + 2 i, x = 1 - 2 i$

Step 2

Find the point at

$x = 1$ .

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Step 2.1

Replace the variable

$x$ with

$1$ in the expression.

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

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

One to any power is one.

$f (1) = \ln (1 - 2 \cdot 1 + 5)$

Step 2.2.1.2

Multiply

$- 2$ by

$1$ .

$f (1) = \ln (1 - 2 + 5)$

Step 2.2.2

Simplify by adding and subtracting.

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Step 2.2.2.1

Subtract

$2$ from

$1$ .

$f (1) = \ln (- 1 + 5)$

Step 2.2.2.2

Add

$- 1$ and

$5$ .

$f (1) = \ln (4)$

Step 2.2.3

The final answer is

$\ln (4)$ .

$\ln (4)$

Step 2.3

Convert

$\ln (4)$ to decimal.

$y = 1.38629436$

Step 3

Find the point at

$x = 2$ .

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Step 3.1

Replace the variable

$x$ with

$2$ in the expression.

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

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

$2$ to the power of

$2$ .

$f (2) = \ln (4 - 2 \cdot 2 + 5)$

Step 3.2.1.2

Multiply

$- 2$ by

$2$ .

$f (2) = \ln (4 - 4 + 5)$

Step 3.2.2

Simplify by adding and subtracting.

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Step 3.2.2.1

Subtract

$4$ from

$4$ .

$f (2) = \ln (0 + 5)$

Step 3.2.2.2

Add

$0$ and

$5$ .

$f (2) = \ln (5)$

Step 3.2.3

The final answer is

$\ln (5)$ .

$\ln (5)$

Step 3.3

Convert

$\ln (5)$ to decimal.

$y = 1.60943791$

Step 4

Find the point at

$x = 3$ .

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Step 4.1

Replace the variable

$x$ with

$3$ in the expression.

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

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

$3$ to the power of

$2$ .

$f (3) = \ln (9 - 2 \cdot 3 + 5)$

Step 4.2.1.2

Multiply

$- 2$ by

$3$ .

$f (3) = \ln (9 - 6 + 5)$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Subtract

$6$ from

$9$ .

$f (3) = \ln (3 + 5)$

Step 4.2.2.2

Add

$3$ and

$5$ .

$f (3) = \ln (8)$

Step 4.2.3

The final answer is

$\ln (8)$ .

$\ln (8)$

Step 4.3

Convert

$\ln (8)$ to decimal.

$y = 2.07944154$

Step 5

The log function can be graphed using the vertical asymptote at

$x = 1 + 2 i, x = 1 - 2 i$ and the points

$(1, 1.38629436), (2, 1.60943791), (3, 2.07944154)$ .

Vertical Asymptote:

$x = 1 + 2 i, x = 1 - 2 i$

$\begin{array}{c} x & y \\ 1 & 1.386 \\ 2 & 1.609 \\ 3 & 2.079 \end{array}$

Step 6