Problem 17
Question
Sketch the graphs of the three functions by hand on the same rectangular coordinate system. Verify your results with a graphing utility.$$\begin{aligned}&f(x)=\frac{1}{x}\\\&g(x)=\frac{1}{x}-2\\\&h(x)=\frac{1}{x-1}+2\end{aligned}$$.
Step-by-Step Solution
Verified Answer
The graph of \(f(x) = 1/x\) has two curved lines in 1st and 2nd quadrants with asymptotes at x=0 and y=0. \(g(x) = 1/x - 2\) is similar but shifted down by 2 units, with asymptote at y=-2. \(h(x) = 1/(x-1) + 2\) is the same shape, but shifted right by 1 unit (asymptote at x=1) and upward by 2 units (asymptote at y=2). These results are confirmed by a graphing tool.
1Step 1: Analyze the Functions
Before sketching, notice how each function is derived from the function \(f(x) = 1/x\), a hyperbola, and what feature is different for each. \(g(x) = 1/x - 2\) is a vertical shift down by 2 units from \(f(x)\) and \(h(x) = 1/(x-1) + 2\) is a horizontal shift to right by 1 unit and an upward shift of 2 units from \(f(x)\).
2Step 2: Sketch the Graph of \(f(x)\)
Draw the Cartesian plane. For \(f(x) = 1/x\), sketch two separate curved lines in the second (x<0) and first (x>0) quadrants which are shaped like half a 'U' lying on its side, with an asymptote at x=0 and y=0.
3Step 3: Sketch the Graph of \(g(x)\)
Next, draw the function \(g(x) = 1/x - 2\) with the same shape as \(f(x)\), but shifted down by 2 units. The asymptote lies on y=-2 instead of y=0.
4Step 4: Sketch the Graph of \(h(x)\)
Finally, for \(h(x) = 1/(x-1) + 2\), the graph will be similar to \(f(x)\), but shifted to the right by one unit (with vertical asymptote at x=1) and upwards by 2 units (horizontal asymptote at y=2).
5Step 5: Verify with a Graphing Tool
Compare your hand-drawn graphs to those produced by a graphing tool to validate your drawings. The shape and position of each graph should match the corresponding graph produced by the tool.
Key Concepts
HyperbolaVertical AsymptoteHorizontal ShiftGraphing Utility
Hyperbola
The function \(f(x) = \frac{1}{x}\) defines a classic hyperbola. Hyperbolas are characterized by two separate curves that lie on opposite quadrants. For the function \(f(x)\), these curves lie in the first and third quadrants, assuming positive values on one and negative on the other. Hyperbolas have two asymptotes, usually the x-axis and y-axis when centered at the origin. These curves will never touch the axes but will get closer and closer indefinitely. As x approaches zero from the positive or negative side, the function values grow towards infinity or negative infinity, respectively. The unique shape and behavior around its asymptotes are what define this function as a hyperbola.
Vertical Asymptote
Vertical asymptotes occur in rational functions where the denominator is zero. In our base function \(f(x) = \frac{1}{x}\), the vertical asymptote is at \(x = 0\). This happens because division by zero is undefined, causing the function to spike to positive or negative infinity near this vertical line.
For the altered functions, the vertical asymptote moves depending on shifts. For \(g(x) = \frac{1}{x} - 2\), the vertical asymptote remains at \(x = 0\), but for \(h(x) = \frac{1}{x-1} + 2\), it's moved to \(x = 1\). This shift is due to the change in the denominator of the function, illustrating how the input can change the position of the asymptote.
For the altered functions, the vertical asymptote moves depending on shifts. For \(g(x) = \frac{1}{x} - 2\), the vertical asymptote remains at \(x = 0\), but for \(h(x) = \frac{1}{x-1} + 2\), it's moved to \(x = 1\). This shift is due to the change in the denominator of the function, illustrating how the input can change the position of the asymptote.
Horizontal Shift
Horizontal shifts in graphing change the horizontal position of the graph without affecting its shape. They result from changes within the function's formula that affect the input variable. For instance, in \(h(x) = \frac{1}{x-1} + 2\), the expression \(x - 1\) inside the denominator indicates a shift to the right by 1 unit.
So, instead of the vertical asymptote lying on \(x = 0\) as it does for \(f(x)\), it moves to \(x = 1\). It’s helpful to remember: if you subtract from \(x\), the graph shifts to the right; if you add to \(x\), it shifts to the left. This concept helps decode the effect of the equation on the graph's position.
So, instead of the vertical asymptote lying on \(x = 0\) as it does for \(f(x)\), it moves to \(x = 1\). It’s helpful to remember: if you subtract from \(x\), the graph shifts to the right; if you add to \(x\), it shifts to the left. This concept helps decode the effect of the equation on the graph's position.
Graphing Utility
Graphing utilities are excellent tools to verify and compare hand-drawn sketches with precise computer-generated graphs.
Modern graphing calculators and software offer ways to enter equations like \(f(x) = \frac{1}{x}\), \(g(x) = \frac{1}{x} - 2\), and \(h(x) = \frac{1}{x-1} + 2\). They accurately depict transformations and allow users to see shifts and asymptotes clearly. This visual confirmation ensures the hand-drawn graphs are correct and helps students understand how algebraic modifications translate to visual changes. Using such utilities enhances learning by allowing for prompt feedback and better comprehension of abstract mathematical concepts.
Modern graphing calculators and software offer ways to enter equations like \(f(x) = \frac{1}{x}\), \(g(x) = \frac{1}{x} - 2\), and \(h(x) = \frac{1}{x-1} + 2\). They accurately depict transformations and allow users to see shifts and asymptotes clearly. This visual confirmation ensures the hand-drawn graphs are correct and helps students understand how algebraic modifications translate to visual changes. Using such utilities enhances learning by allowing for prompt feedback and better comprehension of abstract mathematical concepts.
Other exercises in this chapter
Problem 16
Which sets of ordered pairs represent functions from \(A\) to \(B\) ? Explain. \(A=\\{a, b, c\\}\) and \(B=\\{0,1,2,3\\}\) (a) \(\\{(a, 1),(c, 2),(c, 3),(b, 3)\
View solution Problem 16
Find the slope of the line passing through the pair of points. Then use a graphing utility to plot the points and use the draw feature to graph the line segment
View solution Problem 17
Find (a) \((f+g)(x),\) (b) \((f-g)(x)\) , (c) \((f g)(x),\) and \((d)(f / g)(x) .\) What is the domain of \(f / g ?\) $$f(x)=\frac{1}{x}, \quad g(x)=\frac{1}{x^
View solution Problem 17
Use the graph of the function to answer the questions. (a) Determine the domain of the function. (b) Determine the range of the function. (c) Find the value(s)
View solution