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Chapter 8: Problem 12

If \(f(x)=\frac{1}{x}\) a. Describe the transformations of \(f(x)\) used to create the new functions \(g(x), h(x)\), and \(k(x)\). \(g(x)=\frac{1}{3 x} \quad h(x)=-\frac{2}{x-3} \quad k(x)=\frac{1}{2-x}+4\) b. Determine the domain of each function in part (a). c. Determine the equation of the vertical asymptote for each function in part (a).

Short Answer

Expert verified
Transformations: g(x) horizontal stretch, h(x) vertical stretch and horizontal shift, k(x) reflection and shifts. Domains: \( x eq 0, x eq 3, x eq 2 \). Vertical asymptotes: \( x = 0, 3, 2 \).

Step by step solution

01

- Identify transformation of g(x)

The function \( g(x) = \frac{1}{3x} \) is a transformation of \( f(x) = \frac{1}{x} \) involving a horizontal stretch. The factor of 3 in the denominator stretches the graph horizontally by a factor of 3.
02

- Identify transformation of h(x)

The function \( h(x) = -\frac{2}{x-3} \) involves multiple transformations: a vertical stretch by a factor of 2, a reflection across the x-axis (due to the negative sign), and a horizontal shift 3 units to the right.
03

- Identify transformation of k(x)

The function \( k(x) = \frac{1}{2-x} + 4 \) can be rewritten as \( k(x) = -\frac{1}{x-2} + 4 \). This function involves a horizontal reflection across the axis, a horizontal shift 2 units to the right, and a vertical shift up by 4 units.
04

- Determine the domain of g(x)

For \( g(x) = \frac{1}{3x} \), the denominator cannot be zero. Therefore, the domain is all real numbers except 0. The domain is \( x eq 0 \).
05

- Determine the domain of h(x)

For \( h(x) = -\frac{2}{x-3} \), the denominator cannot be zero. Therefore, the domain is all real numbers except 3. The domain is \( x eq 3 \).
06

- Determine the domain of k(x)

For \( k(x) = \frac{1}{2-x} + 4 \), the denominator cannot be zero. Therefore, the domain is all real numbers except 2. The domain is \( x eq 2 \).
07

- Determine vertical asymptote of g(x)

For \( g(x) = \frac{1}{3x} \), the vertical asymptote occurs where the denominator is zero. The vertical asymptote is \( x = 0 \).
08

- Determine vertical asymptote of h(x)

For \( h(x) = -\frac{2}{x-3} \), the vertical asymptote occurs where the denominator is zero. The vertical asymptote is \( x = 3 \).
09

- Determine vertical asymptote of k(x)

For \( k(x) = \frac{1}{2-x} + 4 \), the vertical asymptote occurs where the denominator is zero. The vertical asymptote is \( x = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Stretch
A horizontal stretch transformation occurs when we multiply the variable inside the function by a factor. For instance, if we have the function \(f(x) = \frac{1}{x}\) and modify it to \(g(x) = \frac{1}{3x}\), we have stretched it horizontally. The factor of 3 causes the graph to expand along the horizontal axis by a factor of 3. This means that the graph will appear wider, as each x-value is divided by 3 before applying the original function.
Vertical Reflection
A vertical reflection occurs when we multiply the entire function by -1. This flips the graph of the function across the x-axis. Taking the function \(f(x) = \frac{1}{x}\) and transforming it to \(h(x) = -\frac{2}{x-3}\) not only reflects it vertically due to the negative sign but also involves a vertical stretch by a factor of 2. In graphical terms, if you took each point on the graph and reflected it over the x-axis, you'd achieve the vertical reflection.
Horizontal Shift
Shifting a function horizontally involves adding or subtracting a constant value from the variable inside the function. For example, consider the function \(h(x) = -\frac{2}{x-3}\). Here, the subtraction of 3 from x inside the function indicates a shift to the right by 3 units. This means every point on the original graph of \(f(x) = \frac{1}{x}\) moves 3 units to the right. If we encounter an expression like \(k(x) = \frac{1}{2-x}\), rewriting it gives us \(k(x) = -\frac{1}{x-2}\), which indicates a shift to the right by 2 units due to \(x-2\).
Vertical Asymptote
Vertical asymptotes occur at values of x that make the denominator of a rational function zero. These lines represent values that the function cannot take. For the function \(f(x) = \frac{1}{x}\), the vertical asymptote is at \(x = 0\) because the denominator is zero at this point. Similarly, for \(g(x) = \frac{1}{3x}\), the vertical asymptote remains at \(x = 0\). For functions like \(h(x) = -\frac{2}{x-3}\), the vertical asymptote occurs at \(x = 3\), and for \(k(x) = \frac{1}{2-x}\), it occurs at \(x = 2\).
Domain of Functions
The domain of a function includes all allowable x-values for which the function is defined. For rational functions, the domain typically excludes values that make the denominator zero. For instance, \(f(x) = \frac{1}{x}\) excludes \(x = 0\), so its domain is \(x eq 0\). Applying this logic: for \(g(x) = \frac{1}{3x}\), the domain is also \(x eq 0\). For \(h(x) = -\frac{2}{x-3}\), x cannot be 3, giving a domain of \(x eq 3\). Lastly, for \(k(x) = \frac{1}{2-x} + 4\), the domain excludes \(x = 2\), making it \(x eq 2\).

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Most popular questions from this chapter

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