Problem 72
Question
Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$r(x)=\frac{2 x^{3}+2 x}{x^{2}-1}$$
Step-by-Step Solution
Verified Answer
Vertical asymptotes at \(x = 1\) and \(x = -1\); slant asymptote at \(y = 2x\). Graph accordingly.
1Step 1: Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero, provided the numerator is not also zero at those points. Set the denominator equal to zero and solve for \(x\): \(x^2 - 1 = 0\). Factoring, we have \((x+1)(x-1)=0\). Thus, the vertical asymptotes occur at \(x=1\) and \(x=-1\).
2Step 2: Determine the Slant Asymptote
A slant (or oblique) asymptote occurs when the degree of the numerator is exactly one more than the degree of the denominator. Here, the numerator, \(2x^3 + 2x\), is degree 3 and the denominator, \(x^2-1\), is degree 2. Perform polynomial long division of \(2x^3 + 0x^2 + 2x\) by \(x^2 - 1\) to find the slant asymptote. Dividing, the quotient is \(2x\), so the slant asymptote is \(y = 2x\).
3Step 3: Sketch the Graph
With the vertical asymptotes at \(x=1\) and \(x=-1\), and the slant asymptote at \(y=2x\), you can sketch the graph. For more accuracy, consider additional properties such as intercepts: the y-intercept is found by evaluating the function at \(x=0\), which gives \(r(0) = 0\). Then, choose several points around the asymptotes to see how the function behaves as \(x\) approaches these lines.
Key Concepts
Understanding Vertical AsymptotesUsing Polynomial Long Division for Slant AsymptotesGraph Sketching TipsAnalyzing Function Behavior
Understanding Vertical Asymptotes
Vertical asymptotes are lines where the function tends to infinity. They occur when the denominator of a rational function equals zero, but not the numerator. This is because division by zero is undefined, causing the function's value to spike upwards or downwards.
To find vertical asymptotes in the function \( r(x) = \frac{2x^3 + 2x}{x^2 - 1} \), we set the denominator \( x^2 - 1 \) equal to zero.
Solving \((x+1)(x-1)=0\) gives \(x=1\) and \(x=-1\). These are where the function will have vertical asymptotes.
Always remember that if the numerator was also zero at these points, it would not be a vertical asymptote.
To find vertical asymptotes in the function \( r(x) = \frac{2x^3 + 2x}{x^2 - 1} \), we set the denominator \( x^2 - 1 \) equal to zero.
Solving \((x+1)(x-1)=0\) gives \(x=1\) and \(x=-1\). These are where the function will have vertical asymptotes.
Always remember that if the numerator was also zero at these points, it would not be a vertical asymptote.
Using Polynomial Long Division for Slant Asymptotes
A slant (or oblique) asymptote appears when the degree of the numerator is one higher than that of the denominator. It is a straight line that the curve approaches as \(x\) goes to infinity.
In our example, \(r(x) = \frac{2x^3 + 2x}{x^2 - 1}\), the numerator is of degree 3 and the denominator is degree 2, so a slant asymptote exists.
To find it, use polynomial long division of \(2x^3 + 2x\) by \(x^2 - 1\).
When dividing, the quotient \(2x\) gives us the slant asymptote \(y = 2x\).
Remember to ignore any remainders; they only affect the curve close to finite values of \(x\).
In our example, \(r(x) = \frac{2x^3 + 2x}{x^2 - 1}\), the numerator is of degree 3 and the denominator is degree 2, so a slant asymptote exists.
To find it, use polynomial long division of \(2x^3 + 2x\) by \(x^2 - 1\).
When dividing, the quotient \(2x\) gives us the slant asymptote \(y = 2x\).
Remember to ignore any remainders; they only affect the curve close to finite values of \(x\).
Graph Sketching Tips
Sketching a graph involves understanding how a function behaves near its asymptotes and intercepts. It's like a roadmap for the function's path.
For \(r(x)\) as provided, we know:
For \(r(x)\) as provided, we know:
- Vertical asymptotes at \(x = 1\) and \(x = -1\).
- A slant asymptote at \(y = 2x\).
- Plot the y-intercept, found by setting \(x = 0\), resulting in \(r(0) = 0\).
- Choose points on either side of the vertical asymptotes \(x = 1\) and \(x = -1\) to see how the function behaves.
- Watch the graph approach \(y = 2x\) as \(x\) heads towards positive or negative infinity.
Analyzing Function Behavior
Function behavior analysis provides insights into how the function acts at important points and asymptotes. The goal is to predict the curve's path and shape.
For \(r(x)\), consider:
Remember, the change in curve near asymptotes can vary from vertical rises to gentle slopes, depending on adjacent function points.
For \(r(x)\), consider:
- The trend around asymptotes as the curve goes to \(+\infty\) or \(-\infty\).
- How the presence of \(y = 2x\) influences the slope at large values of \(x\).
Remember, the change in curve near asymptotes can vary from vertical rises to gentle slopes, depending on adjacent function points.
Other exercises in this chapter
Problem 71
Graph the family of polynomials in the same viewing rectangle, using the given values of \(c .\) Explain how changing the value of \(c\) affects the graph. $$P(
View solution Problem 72
Find all solutions of the equation and express them in the form \(a+b i\) $$x^{2}+\frac{1}{2} x+1=0$$
View solution Problem 72
So far, we have worked only with polynomials that have real coefficients. These exercises involve polynomials with real and imaginary coefficients. (a) Find the
View solution Problem 72
Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$P(x)=x^{4}-2 x^{3}-9 x^{2}+2 x+8 ; \quad a=-3,
View solution