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Chapter 4: Problem 40

Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to confirm your answer. \(s(x)=\frac{x+2}{(x+3)(x-1)}\)

Short Answer

Expert verified
X-intercept: (-2, 0), y-intercept: (0, -2/3); vertical asymptotes at x = -3 and x = 1; horizontal asymptote at y = 0.

Step by step solution

01

Finding the x-intercepts

To find the x-intercepts of the function, set the numerator equal to zero: \[x + 2 = 0\]Solve for \(x\): \[x = -2\]The x-intercept is at \((-2, 0)\).
02

Finding the y-intercepts

To find the y-intercept, evaluate the function at \(x = 0\): \[s(0) = \frac{0 + 2}{(0 + 3)(0 - 1)} = \frac{2}{-3} = -\frac{2}{3}\]The y-intercept is at \((0, -\frac{2}{3})\).
03

Finding vertical asymptotes

Vertical asymptotes occur where the denominator equals zero, given the numerator is not zero at those points. Set the denominator equal to zero: \[(x+3)(x-1) = 0\]This gives:- \(x + 3 = 0\), resulting in \(x = -3\)- \(x - 1 = 0\), resulting in \(x = 1\)The vertical asymptotes are at \(x = -3\) and \(x = 1\).
04

Finding the horizontal asymptote

For rational functions of the form \(\frac{a_n x^n + \, \ldots \,}{b_m x^m + \, \ldots \,}\), if \(n < m\), the horizontal asymptote is \(y = 0\). Here, the degree of the numerator (1) is less than the denominator (2), so the horizontal asymptote is \(y = 0\).
05

Sketching and confirming the graph

We identified the x-intercept at \((-2, 0)\), the y-intercept at \((0, -\frac{2}{3})\), vertical asymptotes at \(x = -3\) and \(x = 1\), and the horizontal asymptote at \(y = 0\). Plot these features on a coordinate plane and sketch the graph around the vertical asymptotes, showing it approaches the horizontal asymptote. Use a graphing tool to confirm the graph's shape and features.

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Key Concepts

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

x-intercepts: Where the Graph Crosses the x-axis
In a rational function like \(s(x)=\frac{x+2}{(x+3)(x-1)}\), x-intercepts are the points where the graph crosses the x-axis. This happens when the function value equals zero. To find the x-intercepts, we focus on the numerator because a fraction is zero when the numerator is zero (and the denominator isn't infinite). So, we set \(x+2 = 0\) and solve for \(x\):
  • \(x = -2\)
This means the graph crosses the x-axis at the point \((-2, 0)\). Remember, x-intercepts are always in the form \((x,0)\) because they sit right on the x-axis.
By understanding that solving \(x+2 = 0\) gives us the x-intercepts, you can now find x-intercepts in other rational functions by performing a similar procedure.
y-intercepts: Where the Graph Meets the y-axis
To find the y-intercepts of a rational function like \(s(x)=\frac{x+2}{(x+3)(x-1)}\), we determine where the graph crosses the y-axis. This occurs when \(x = 0\), as every point on the y-axis has an \(x\)-coordinate of zero. Plugging \(x = 0\) into the rational function:
  • \(s(0) = \frac{0+2}{(0+3)(0-1)} = \frac{2}{-3} = -\frac{2}{3}\)
This gives us the y-intercept at \((0, -\frac{2}{3})\). It's worth noting that y-intercepts have the form \((0, y)\), indicating how high or low they land on the y-axis. By evaluating the function at \(x=0\), you can find y-intercepts for other rational functions as well.
Vertical Asymptotes: The Lines the Graph Can Never Touch
Vertical asymptotes are vertical lines on a graph where the function becomes undefined. These occur where the denominator of the rational function equals zero, provided the numerator is not zero at those points. For our function \(s(x)=\frac{x+2}{(x+3)(x-1)}\), identifying these points involves setting the entire denominator equal to zero:
  • \((x+3)(x-1) = 0\)
  • This results in two separate equations: \(x+3 = 0\) and \(x-1 = 0\)
  • Solving these, we find \(x = -3\) and \(x = 1\)
Thus, the vertical asymptotes are at \(x = -3\) and \(x = 1\). Since the function approaches these lines but never crosses them, vertical asymptotes act as boundaries that help shape the overall graph's behavior. In general, solving for zeros in the denominator will reveal vertical asymptotes for other rational functions too.
Horizontal Asymptotes: The Ultimate Direction of the Graph
Horizontal asymptotes reveal the behavior of a rational function as \(x\) approaches infinity or negative infinity. Generally, they show us the value the function is tending towards as it goes off the chart. For a rational function such as \(s(x)=\frac{x+2}{(x+3)(x-1)}\), the horizontal asymptote depends on the degrees of the numerator and denominator:
  • The degree of the numerator is 1 (since \(x\) is the highest power in \(x+2\))
  • The degree of the denominator is 2 (as from \((x+3)(x-1)\), \(x^2\) is the highest power)
  • Since 1 < 2, according to the rules of rational functions, there is a horizontal asymptote at \(y = 0\)
Keep in mind, this rule applies broadly: if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\). This knowledge can guide you in finding horizontal asymptotes of other rational functions as well.

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Most popular questions from this chapter

Volume of a Silo A grain silo consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is \(15,000 \mathrm{ft}^{3}\) and the cylindrical part is 30 \(\mathrm{ft}\) tall, what is the radius of the silo, correct to the nearest tenth of a foot?

By the Zeros Theorem, every \(n\) th-degree polynomial equation has exactly \(n\) solutions (including possibly some that are repeated). Some of these may be real and some may be imaginary. Use a graphing device to determine how many real and imaginary solutions each equation has. (a) \(x^{4}-2 x^{3}-11 x^{2}+12 x=0\) (b) \(x^{4}-2 x^{3}-11 x^{2}+12 x-5=0\) (c) \(x^{4}-2 x^{3}-11 x^{2}+12 x+40=0\)

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