Problem 19
Question
Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph.) $$ f(t)=\frac{t^{2}}{t^{2}-9} $$
Step-by-Step Solution
Verified Answer
The function \(f(t)=\frac{t^{2}}{t^{2}-9}\) has vertical asymptotes at \(t = 3\) and \(t = -3\), and a horizontal asymptote at \(y = 1\).
1Step 1: Identify the value(s) of t that make the denominator to be zero
To find the value(s) of t for which the function's denominator is zero, we need to solve the following equation:
$$
t^2-9=0
$$
2Step 2: Solve the equation to find the vertical asymptotes
We have the quadratic equation \(t^2 - 9 = 0\). This equation can be factored as \((t-3)(t+3) = 0\). This equation has two solutions: \(t=3\) and \(t=-3\). These are the values of t where the denominator becomes zero. Hence, we have:
- Vertical asymptote 1: \(t = 3\)
- Vertical asymptote 2: \(t = -3\)
Now, let's move on to find the horizontal asymptote.
3Step 3: Determine the horizontal asymptote
We can find the horizontal asymptote by finding the limit of the function as t approaches infinity or negative infinity. The function is given as:
$$
f(t)=\frac{t^{2}}{t^{2}-9}
$$
We can determine the behavior of the function for large positive and negative values of t. Let's find the limit as t approaches infinity:
$$
\lim_{t\rightarrow\infty}{\frac{t^{2}}{t^{2}-9}}
$$
In this case, the degree of the numerator and the denominator are equal (both are 2). To find the limit in this case, we can simply take the ratio of the leading coefficients, which are 1 in both cases:
$$
\lim_{t\rightarrow\infty}{\frac{t^{2}}{t^{2}-9}} = \frac{1}{1} = 1
$$
Since the limit is a constant value when t approaches infinity, we can conclude that there is a horizontal asymptote at \(y = 1\).
#Results#:
- Vertical asymptotes: \(t = 3\) and \(t = -3\)
- Horizontal asymptote: \(y = 1\)
Key Concepts
Vertical AsymptotesHorizontal AsymptotesRational Functions
Vertical Asymptotes
Vertical asymptotes occur where the function's denominator equals zero because the function becomes undefined at these points. For the given function, \(f(t) = \frac{t^2}{t^2 - 9}\), finding vertical asymptotes involves solving \(t^2 - 9 = 0\). By factoring, we have \((t - 3)(t + 3) = 0\), meaning \(t = 3\) and \(t = -3\) are solutions.
- When \(t = 3\) or \(t = -3\), the denominator of the function becomes zero, thus the function shoots off to infinity or negative infinity.
- This creates vertical asymptotes, which are essentially lines t = 3 and t = -3 that the graph approaches but never touches.
Horizontal Asymptotes
Horizontal asymptotes describe the behavior of a graph as the input value t (in this function) becomes very large or very small. For our function \(f(t) = \frac{t^2}{t^2 - 9}\), both the numerator and the denominator are quadratic terms.
To find the horizontal asymptote, note that the degrees (highest exponent) of the numerator and denominator are the same (both 2 in this case). This implies that as \(t\) approaches infinity, the fractions' leading coefficients determine the horizontal asymptote, which is a constant value:
To find the horizontal asymptote, note that the degrees (highest exponent) of the numerator and denominator are the same (both 2 in this case). This implies that as \(t\) approaches infinity, the fractions' leading coefficients determine the horizontal asymptote, which is a constant value:
- Leading coefficients for both numerator and denominator are 1, so we have \(\frac{1}{1} = 1\).
- The horizontal asymptote, therefore, is \(y = 1\).
Rational Functions
Rational functions are ratios of two polynomial expressions, where one polynomial is divided by another. In the function \(f(t) = \frac{t^2}{t^2 - 9}\), \(t^2\) is the numerator and \(t^2 - 9\) is the denominator indicating it's a rational function.
These functions often present interesting and predictable properties such as:
These functions often present interesting and predictable properties such as:
- Vertical asymptotes: Locations where the function is undefined due to division by zero, as is seen when \(t = 3\) and \(t = -3\).
- Horizontal asymptotes: Describe how the function behaves at extreme positive or negative values of the input. In our case, the grip tightens towards \(y = 1\).
- Holes: Occur when a factor of the denominator cancels out with a factor in the numerator, which isn't present in this specific function but noteworthy in understanding other rational functions.
Other exercises in this chapter
Problem 19
The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $$\$ 600 /$$ person/ day if exactly 20 people sign up for the cruise. However,
View solution Problem 19
Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x^{3}+3 x^{2}-1 \text { on }[-3,2] $$
View solution Problem 19
Show that the function is concave upward wherever it is defined. $$ f(x)=4 x^{2}-12 x+7 $$
View solution Problem 19
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ g(x)=x^{3}+3 x^{2}+1 $$
View solution