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Chapter 3: Problem 57

Graph the functions by using transformations of the graphs of \(y=\frac{1}{x}\) and \(y=\frac{1}{x^{2}}\). $$ h(x)=\frac{1}{x^{2}}+2 $$

Short Answer

Expert verified
Shift \(f(x)=\frac{1}{x^2}\) up by 2 units to get \(h(x)=\frac{1}{x^2}+2\).

Step by step solution

01

Identify the Base Function

The given function is similar to the base function: \(f(x) = \frac{1}{x^2}\).
02

Vertical Shift

The given function is \(h(x)=\frac{1}{x^2} + 2\). This indicates that the graph of \(f(x)=\frac{1}{x^2}\) has been shifted vertically upward by 2 units.
03

Draw the Base Graph

Start by drawing the graph of \(f(x)=\frac{1}{x^2}\). This graph has a vertical asymptote at \(x=0\) and a horizontal asymptote at \(y=0\). It is also always positive and symmetric about the y-axis.
04

Apply Vertical Shift

Shift the entire graph of \(f(x)=\frac{1}{x^2}\) upward by 2 units. This means that the new horizontal asymptote is at \(y=2\), and every point on the graph of \(f(x)\) moves 2 units up.
05

Plot the Transformed Graph

After the shift, the transformed graph of \(h(x)=\frac{1}{x^2} + 2\) should be plotted. It will have the new horizontal asymptote at \(y=2\) and will be 2 units higher everywhere compared to the graph of \(f(x) = \frac{1}{x^2}\).
06

Verify Key Points

Pick key points from the graph of \(f(x)=\frac{1}{x^2}\) such as \((1,1)\) and \((2, \frac{1}{4})\), and then add 2 to the y-values to plot points on the graph of \(h(x)\) such as \((1,3)\) and \((2, \frac{1}{4} + 2 = 2.25)\).

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Key Concepts

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

Vertical Shift
In graph transformations, a vertical shift adjusts the entire graph up or down along the y-axis. For the function provided, we start with the base function \(f(x) = \frac{1}{x^{2}}\). The transformation we apply is a vertical shift upwards by 2 units to obtain our transformed function, \( h(x)=\frac{1}{x^{2}}+2\). Here is how this impacts the graph:
  • The whole graph of \( f(x) = \frac{1}{x^2}\) moves 2 units up.
  • Every point on the graph will now be higher by 2 units.
  • The new horizontal asymptote will shift from \(y=0\) to \(y=2\).
As a result, the curve remains in the same shape but is elevated by 2 units. This is a simple but powerful transformation that can easily be visualized.
Asymptotes
Asymptotes are lines that a graph approaches but never touches or crosses. For rational functions like \( y=\frac{1}{x^{2}}\), understanding asymptotes is crucial:
  • Vertical Asymptote: This occurs where the function is undefined. For \( y=\frac{1}{x^{2}}\), it happens at \(x=0\). Regardless of transformations, this doesn鈥檛 change since there is no value of \(x\) that will make \( \frac{1}{x^2} \) defined at \( x=0\).
  • Horizontal Asymptote: Initially, for \( y=\frac{1}{x^{2}}\), the horizontal asymptote is at \( y=0\) because as \( x\) goes to infinity, \( \frac{1}{x^2} \) gets closer to 0. With a vertical shift upwards by 2 units, the new horizontal asymptote becomes \( y=2 \).
Knowing how asymptotes will behave under transformations allows us to sketch more accurate graphs.
Base Function
The base function for our example is \( f(x) = \frac{1}{x^{2}} \). Understanding this base function is the foundation for applying transformations:
  • This function is always positive because squaring any real number and taking its reciprocal results in a positive value.
  • The function is symmetric with respect to the y-axis. This means that the graph looks the same on both sides of the y-axis.
  • \( f(x) = \frac{1}{x^{2}} \) has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 0, \) meaning the graph approaches but never touches these lines.
Understanding these properties helps in plotting any transformations accurately. Once you master the base function, transitions such as shifts and stretches become simpler and more intuitive.

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Most popular questions from this chapter

Sketch a rational function subject to the given conditions. Answers may vary. Horizontal asymptote: \(y=3\) Vertical asymptotes: \(x=-1\) and \(x=1\) \(y\) -intercept: (0,0) \(x\) -intercept (0,0) Symmetric to the \(y\) -axis Passes through the point (2,4)

Suppose that \(y\) varies directly as \(x^{5}\) and inversely as \(w^{2}\). If both \(x\) and \(w\) are doubled, what is the effect on \(y ?\)

Sometimes it is necessary to use a "friendly" viewing window on a graphing calculator to see the key features of a graph. For example, for a calculator screen that is 96 pixels wide and 64 pixels high, the "decimal viewing window" defined by [-4.7,4.7,1] by [-3.1,3.1,1] creates a scaling where each pixel represents 0.1 unit. The window [-9.4,9.4,1] by [-6.2,6.2,1] defines each pixel as 0.2 unit, and so on. Exercises \(112-113\) compare the use of the standard viewing window to a "friendly" viewing window. a. Identify any vertical asymptotes of the function defined by \(f(x)=\frac{x^{2}+3 x+2}{x+1}\). b. Compare the graph of \(f(x)=\frac{x^{2}+3 x+2}{x+1}\) on the standard viewing window [-10,10,1] by [-10,10,1 and on the window [-4.7,4.7,1] by [-3.1,3.1,1] . Which graph shows the behavior at \(x=-1\) more completely?

The procedure to solve a polynomial or rational inequality may be applied to all inequalities of the form \(f(x)>0, f(x)<0,\) \(f(x) \geq 0,\) and \(f(x) \leq 0 .\) That is, find the real solutions to the related equation and determine restricted values of \(x .\) Then determine the sign of \(f(x)\) on each interval defined by the boundary points. Use this process to solve the inequalities. $$ \frac{1}{\sqrt{x-3}-5} \leq 0 $$

Explain why \(\frac{x^{2}+2}{x^{2}+1}<0\) has no solution.
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