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Chapter 9: Problem 2

a) Determine the roots of the rational equation \(-\frac{2}{x}+x+1=0\) algebraically. b) Graph the rational function \(y=-\frac{2}{x}+x+1\) and determine the \(x\) -intercepts. c) Explain the connection between the roots of the equation and the \(x\) -intercepts of the graph of the function.

Short Answer

Expert verified
The roots, \(x=1\) and \(x=-2\), are also the x-intercepts of the function's graph.

Step by step solution

01

Move all terms to one side

First, rewrite the equation \(-\frac{2}{x}+x+1=0\) with all terms on one side: \(-\frac{2}{x} + x + 1 = 0 \).
02

Multiply through by the common denominator

Multiply every term by the common denominator, which is \(x\): \(-2 + x^2 + x=0 \).
03

Rearrange into standard quadratic form

Rewrite the equation in standard quadratic form: \(x^2 + x - 2 = 0 \).
04

Solve the quadratic equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) where \(a=1, b=1, c=-2\): \(x = \frac{-1 \pm \sqrt{(1)^2-4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1+8}}{2} = \frac{-1 \pm 3}{2}\). So, \(x=1\) and \(x=-2\).
05

Graph the function

Plot the function \(y = -\frac{2}{x} + x + 1\) on a Cartesian plane. Identify the points where the graph touches or crosses the x-axis. These points are the solutions we found in step 4: \(x=1\) and \(x=-2\).
06

Determine the x-intercepts

From the graph, observe that the x-intercepts are at \(x=1\) and \(x=-2\). These are the points where the function \(y\) equals zero.
07

Explain the connection

The roots of the equation \(-\frac{2}{x} + x + 1 = 0\) are the same as the x-intercepts of the function \(y=-\frac{2}{x}+x+1\). This is because the roots of the equation represent the values of \(x\) where the function equals zero, which corresponds to the x-intercepts on the graph.

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Key Concepts

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

Roots of equations
To solve an equation, you need to find the roots, which are the values of the variable that make the equation true. For the equation \(-\frac{2}{x}+x+1=0\), the roots are the values of \(x\) that satisfy this condition. Think of it as the points where the function touches or crosses the x-axis, making the equation equal to zero. Here, we moved all terms to one side and simplified the equation to \(x^2 + x - 2 = 0\) to solve it more easily.
Quadratic formula
The quadratic formula is a critical tool for solving quadratic equations like \(x^2 + x - 2 = 0\). The formula is \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\). Plugging in our values: \(a=1\), \(b=1\), and \(c=-2\), we get \(x = \frac{-1 \pm \sqrt{1+8}}{2} = \frac{-1 \pm 3}{2}\). This gives us two roots, \(x=1\) and \(x=-2\). These roots are essential for graphing and further analysis.
Graphing rational functions
Graphing the function \(y = -\frac{2}{x} + x + 1\) helps visualize where the function intersects the x-axis. These intersection points correspond to the roots of the equation, meaning where \(y = 0\). When graphing, plot several values and connect the points smoothly. Notice how the graph behaves near the x-intercepts \(x=1\) and \(x=-2\). This visual aid is beneficial for understanding the function's behavior and verifying the algebraic solutions.
x-intercepts
The \(x\)-intercepts of a function are the points where the graph crosses the x-axis. For the function \(y = -\frac{2}{x} + x + 1\), the \(x\)-intercepts are at \(x=1\) and \(x=-2\). These intercepts confirm the roots we found using the quadratic formula. The \(x\)-intercepts represent the values of \(x\) that make the output, or \(y\)-value, zero. Check these points by setting \(y = 0\) and solving the resulting equation to find the same roots.
Solving algebraic equations
Solving algebraic equations involves finding the variable values that satisfy the equation. For our example, after rewriting the equation \(-\frac{2}{x} + x + 1 = 0\), simplify it to a quadratic form, \(x^2 + x - 2 = 0\). Use techniques like the quadratic formula to find solutions. Always simplify and verify your solutions by plugging them back into the original equation. Each root reveals important characteristics about the graph of the corresponding function.

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Most popular questions from this chapter

Consider the functions \(f(x)=\frac{x+a}{x+b}\) \(g(x)=\frac{x+a}{(x+b)(x+c)},\) and \(h(x)=\frac{(x+a)(x+c)}{(x+b)(x+c)},\) where \(a, b,\) and \(c\) are different real numbers. a) Which pair of functions do you think will have graphs that appear to be most similar to each other? Explain your choice. b) What common characteristics will all three graphs have? Give reasons for your answer.

The function \(h(v)=\frac{6378 v^{2}}{125-v^{2}}\) gives the maximum height, \(h,\) in kilometres, as a function of the initial velocity, \(v,\) in kilometres per second, for an object launched upward from Earth's surface, if the object gets no additional propulsion and air resistance is ignored. a) Graph the function. What parts of the graph are applicable to this situation? b) Explain what the graph indicates about how the maximum height is affected by the initial velocity. c) The term escape velocity refers to the initial speed required to break free of a gravitational field. Describe the nature of the graph for its non- permissible value, and explain why it represents the escape velocity for the object.

The student council at a large high school is having a fundraiser for a local charity. The council president suggests that they set a goal of raising \(\$ 4000\) a) Let \(x\) represent the number of students who contribute. Let \(y\) represent the average amount required per student to meet the goal. What function \(y\) in terms of \(x\) represents this situation? b) Graph the function. c) Explain what the behaviour of the function for various values of \(x\) means in this context. d) How would the equation and graph of the function change if the student council also received a \(\$ 1000\) donation from a local business?

A truck leaves Regina and drives eastbound. Due to road construction, the truck takes \(2 \mathrm{h}\) to travel the first \(80 \mathrm{km} .\) Once it leaves the construction zone, the truck travels at \(100 \mathrm{km} / \mathrm{h}\) for the rest of the trip. a) Let \(v\) represent the average speed, in kilometres per hour, over the entire trip and \(t\) represent the time, in hours, since leaving the construction zone. Write an equation for \(v\) as a function of \(t\) b) Graph the function for an appropriate domain. c) What are the equations of the asymptotes in this situation? Do they have meaning in this situation? Explain. d) How long will the truck have to drive before its average speed is \(80 \mathrm{km} / \mathrm{h} ?\) e) Suppose your job is to develop GPS technology. How could you use these types of calculations to help travellers save fuel?

For each function, predict the locations of any vertical asymptotes, points of discontinuity, and intercepts. Then, graph the function to verify your predictions. a) \(y=\frac{x^{2}+4 x}{x^{2}+9 x+20}\) b) \(y=\frac{2 x^{2}-5 x-3}{x^{2}-1}\) c) \(y=\frac{x^{2}+2 x-8}{x^{2}-2 x-8}\) d) \(y=\frac{2 x^{2}+7 x-15}{9-4 x^{2}}\)
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