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Chapter 1: Problem 30

\(23-36=\) Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations. $$y=\frac{2}{x}-2$$

Short Answer

Expert verified
Vertical stretch by 2 and shift down by 2 units applied to \( y = \frac{1}{x} \).

Step by step solution

01

Identify the Standard Function

The function we are working with is \( y = \frac{2}{x} - 2 \). The standard function that forms the basis of our transformation is \( y = \frac{1}{x} \). This function is a rational function known for its hyperbolic shape, with vertical asymptote at \( x=0 \) and horizontal asymptote at \( y=0 \).
02

Apply Vertical Stretch

The given function \( y = \frac{2}{x} \) can be seen as a vertical stretch of the standard function \( y = \frac{1}{x} \). The factor \( 2 \) in the numerator stretches the graph vertically by a factor of 2. This means the values of \( y \) become twice as large for each \( x \) compared to the standard function \( y = \frac{1}{x} \).
03

Apply Vertical Shift

Next, the transformation \( y = \frac{2}{x} - 2 \) from the function \( y = \frac{2}{x} \) is a vertical shift. Subtracting 2 from the whole expression shifts the entire graph downward by 2 units. Consequently, this affects the horizontal asymptote, which is now \( y = -2 \) instead of \( y = 0 \).
04

Sketch the Transformed Function

With the graph of \( y = \frac{1}{x} \) vertically stretched and shifted, we can sketch \( y = \frac{2}{x} - 2 \). The resulting graph is similar in shape to \( y = \frac{1}{x} \), but steeper due to the vertical stretch, and positioned such that its horizontal asymptote is at \( y = -2 \) and its vertical asymptote remains at \( x = 0 \). Make sure to reflect these changes in your hand-drawn graph.

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Key Concepts

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

Vertical Stretch
A vertical stretch is a transformation that changes the vertical scale of a graph. For the function \( y = \frac{2}{x} \), the number \( 2 \) in the numerator implies that each \( y \)-value is scaled by a factor of 2 compared to the function \( y = \frac{1}{x} \). Here's how it works:
  • Imagine grabbing the graph at every point and pulling it away from the horizontal axis, effectively doubling its distance.
  • This does not affect the positions of the asymptotes (the lines that the graph approaches but never touches).
The resulting graph is steeper, and this idea of changing the graph's steepness or spread vertically is what defines a vertical stretch.
Vertical Shift
A vertical shift involves moving the graph up or down along the \( y \)-axis. In our function, \( y = \frac{2}{x} - 2 \), the \(-2\) indicates a downward shift of the entire graph.Here's what happens:
  • Every point of the graph is moved 2 units downward. This affects the horizontal asymptote, shifting it from \( y = 0 \) to \( y = -2 \).
  • The vertical stretch from the earlier step remains intact, meaning the graph stays steeper than the standard function.
Therefore, vertical shifts are additive transformations that affect the baseline or starting point of the graph without altering its shape.
Asymptotes
An asymptote is a line that a graph approaches but never actually touches. Understanding asymptotes is key to working with rational functions like \( y = \frac{2}{x} - 2 \).For this graph, we have:
  • A vertical asymptote at \( x = 0 \), which remains unchanged regardless of stretching or shifting. It indicates that the graph will go towards infinity as \( x \) approaches 0 from either direction.
  • A horizontal asymptote was originally at \( y = 0 \) for the standard function \( y = \frac{1}{x} \). However, due to the vertical shift, it moves to \( y = -2 \).
Asymptotes serve as guides, showing where the graph levels out or remains undefined.
Transformations of Functions
Transformations involve altering a function's graph in a predictable manner. When handling rational functions like \( y = \frac{2}{x} - 2 \), the key transformations involve stretching and shifting.To transform a function's graph:
  • First, identify the changes in scale, which involve stretches or compressions. In our case, this is the vertical stretch by factor 2.
  • Next, check for shifts. For our function, we have a vertical shift moving the graph 2 units down.
Understanding these transformations allows you to graph complex rational functions by hand, starting from a basic graph like \( y = \frac{1}{x} \) and applying these modifications accordingly.

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Most popular questions from this chapter

Given that $$\lim _{x \rightarrow 2} f(x)=4 \quad \lim _{x \rightarrow 2} g(x)=-2 \quad \lim _{x \rightarrow 2} h(x)=0$$ find the limits that exist. If the limit does not explain why. (a) $$\lim _{x \rightarrow 2}[f(x)+5 g(x)]$$ (b) $$\lim _{x \rightarrow 2}[g(x)]^{3}$$ (c) $$\lim _{x \rightarrow 2} \sqrt{f(x)}$$ (d) $$\lim _{x \rightarrow 2} \frac{3 f(x)}{g(x)}$$ (e) $$\lim _{x \rightarrow 2} \frac{g(x)}{h(x)}$$ (f) $$\lim _{x \rightarrow 2} \frac{g(x) h(x)}{f(x)}$$

Find the limit. $$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta+\tan \theta}$$

Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{|x|}\right)$$

(a) Show that $$\lim _{x \rightarrow \infty} \frac{4 x^{2}-5 x}{2 x^{2}+1}=2$$ (b) By graphing the function in part (a) and the line \(y=1.9\) on a common screen, find a number \(N\) such that $$\text {if} \quad x>N \quad \text { then } \quad \frac{4 x^{2}-5 x}{2 x^{2}+1}>1.9$$ What if 1.9 is replaced by 1.99\(?\)

(a) A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the tank at a rate of 25 \(\mathrm{L} / \mathrm{min}\) . Show that the concentration of salt \(t\) minutes later (in grams per liter) is $$C(t)=\frac{30 t}{200+t}$$ (b) What happens to the concentration as \(t \rightarrow \infty ?\)
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