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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblem 88
Chapter 2, Problem 88

In Exercises 85–90, find the x-intercepts of the graph of each equation. Then use the x-intercepts to match the equation with its graph. [The graphs are labeled (a) through (f).]
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y=x−2−x−1−6y = x^{-2} - x^{-1} - 6

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To find the x-intercepts of the graph, set the equation equal to zero: \(y = x^{-2} - x^{-1} - 6 = 0\).
Rewrite the negative exponents as fractions to make the equation easier to work with: \(\frac{1}{x^2} - \frac{1}{x} - 6 = 0\).
Multiply both sides of the equation by \(x^2\) (assuming \(x \neq 0\)) to eliminate the denominators: \(1 - x - 6x^2 = 0\).
Rearrange the equation into standard quadratic form: \(-6x^2 - x + 1 = 0\). You can multiply both sides by \(-1\) to get \(6x^2 + x - 1 = 0\) for easier solving.
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=6\), \(b=1\), and \(c=-1\) to find the values of \(x\) where the graph crosses the x-axis.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

X-Intercepts of a Graph

X-intercepts are the points where the graph crosses the x-axis, meaning the y-value is zero. To find them, set y = 0 and solve the resulting equation for x. These points help identify key features of the graph and are essential for matching equations to their graphs.
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Solving Equations with Negative Exponents

Negative exponents indicate reciprocals, such as x^(-1) = 1/x. When solving equations with negative exponents, rewrite terms as fractions to simplify and solve. Understanding this helps in manipulating the equation to find x-intercepts accurately.
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Graph Behavior of Rational Functions

Functions with negative exponents often represent rational functions with variables in denominators. Their graphs can have vertical asymptotes and discontinuities where the denominator is zero. Recognizing these behaviors aids in matching equations to their correct graphs.
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Related Practice
Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. 3x2=603x^2 = 60

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Textbook Question

Solve each radical equation in Exercises 88–89. √ (2x-3) + x = 3

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Textbook Question

The equations in Exercises 79–90 combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 4/(x - 2) + 3/(x + 5) = 7/(x + 5)(x - 2)

Textbook Question

In Exercises 59–94, solve each absolute value inequality. 5 > |4 - x|

Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. x2−2x=1x^2 - 2x = 1

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Textbook Question

In Exercises 59–94, solve each absolute value inequality. 1 < |2 - 3x|

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