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Question 1165857: Given the function f(x)=1/x
a) Write the equation g(x) after the following transformations:
horizontal compression by the factor 1/5 ,
vertical stretch by the factor 7
reflection in the y-axis
translation 10 units left and 1 unit down
b) State the domain and range of g(x)
Steps needed also
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! Here's how to determine the equation, domain, and range of $g(x)$ after the given transformations.
The original function is $f(x) = \frac{1}{x}$.
## a. Write the Equation $g(x)$
A transformation of a function $f(x)$ can be expressed in the form $g(x) = a \cdot f(b(x-h)) + k$, where:
* $|a|$ is the **vertical stretch/compression** factor. (A negative $a$ implies a reflection across the x-axis.)
* $|1/b|$ is the **horizontal stretch/compression** factor. (A negative $b$ implies a reflection across the y-axis.)
* $h$ is the **horizontal shift** (translation).
* $k$ is the **vertical shift** (translation).
### Step 1: Apply Horizontal Compression and Reflection in the $y$-axis
* **Horizontal compression by factor $1/5$** means $b = 5$.
* **Reflection in the $y$-axis** means $b$ is negative, so $b = -5$.
$$f(x) \to f(-5x) = \frac{1}{-5x}$$
### Step 2: Apply Vertical Stretch
* **Vertical stretch by factor 7** means $a = 7$.
$$f(-5x) \to 7f(-5x) = 7 \left(\frac{1}{-5x}\right) = -\frac{7}{5x}$$
### Step 3: Apply Translations
* **Translation 10 units left** means $h = -10$. We replace $x$ with $(x - (-10)) = (x+10)$.
* **Translation 1 unit down** means $k = -1$.
Substitute $(x+10)$ for $x$ and add $k=-1$:
$$g(x) = -\frac{7}{5(x+10)} - 1$$
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## The Final Equation $g(x)$
$$g(x) = -\frac{7}{5(x+10)} - 1$$
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## b. State the Domain and Range of $g(x)$
The domain and range of a transformed function $g(x) = a \cdot f(b(x-h)) + k$ relate directly to the transformations applied to the parent function $f(x)$.
The parent function $f(x) = \frac{1}{x}$ has:
* Domain: $x \neq 0$ (Vertical Asymptote at $x=0$)
* Range: $y \neq 0$ (Horizontal Asymptote at $y=0$)
### Domain of $g(x)$
The domain is restricted by the **vertical asymptote**, which is shifted horizontally.
The vertical asymptote of $f(x)$ is $x=0$.
The only horizontal shift is **10 units left** ($h=-10$).
* New Vertical Asymptote: $x = 0 - 10 = -10$.
* The domain is all real numbers except $x=-10$.
$$\text{Domain of } g(x): \mathbf{\{x \in \mathbb{R} \mid x \neq -10\} \text{ or } (-\infty, -10) \cup (-10, \infty)}$$
### Range of $g(x)$
The range is restricted by the **horizontal asymptote**, which is shifted vertically.
The horizontal asymptote of $f(x)$ is $y=0$.
The only vertical shift is **1 unit down** ($k=-1$).
* New Horizontal Asymptote: $y = 0 - 1 = -1$.
* The range is all real numbers except $y=-1$.
$$\text{Range of } g(x): \mathbf{\{y \in \mathbb{R} \mid y \neq -1\} \text{ or } (-\infty, -1) \cup (-1, \infty)}$$
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