Problem 4 - Algebra and Trigonometry

Question

A rational function is given. (a) Complete each table for the function. (b) Describe the behavior of the function nea its vertical asymptote, based on Tables 1 and \(2 .\) (c) Determine the horizontal asymptote, based on Tables 3 and \(4 .\) \(\begin{array}{|c|c|}\hline x & {r(x)} \\ \hline 1.5 & {} \\ {1.9} & {} \\\ {1.99} & {} & {} \\ {1.999} & {} \\ \hline\end{array}\) \(\begin{array}{|c|c|}\hline x & {r(x)} \\ \hline 2.5 & {} \\ {2.1} & {} & {} \\\ {2.01} & {} \\ {2.001} & {} \\ \hline\end{array}\) \(\begin{array}{|c|c|}\hline x & {r(x)} \\ \hline 10 & {} \\ {50} & {} \\\ {100} & {} \\ {1000} & {} \\ \hline\end{array}\) \(\begin{array}{|c|c|}\hline x & {r(x)} \\ \hline-10 & {} \\ {-50} & {} \\ {-100} & {} \\ {-1000} & {} \\\ \hline\end{array}\) \(r(x)=\frac{3 x^{2}+1}{(x-2)^{2}}\)

Step-by-Step Solution

Answer

Vertical asymptote at \(x = 2\), horizontal asymptote at \(y = 3\).

Step 1: Evaluate Table 1

Substitute each value of \(x\) in Table 1 into the function \(r(x) = \frac{3x^2 + 1}{(x-2)^2}\) to find \(r(x)\).- For \(x = 1.5\): \[ r(1.5) = \frac{3(1.5)^2 + 1}{(1.5 - 2)^2} = \frac{3(2.25) + 1}{(0.5)^2} = \frac{6.75 + 1}{0.25} = \frac{7.75}{0.25} = 31\]- For \(x = 1.9\): \[ r(1.9) = \frac{3(1.9)^2 + 1}{(1.9 - 2)^2} = \frac{3(3.61) + 1}{(0.1)^2} = \frac{10.83 + 1}{0.01} = \frac{11.83}{0.01} = 1183\]- For \(x = 1.99\): \[ r(1.99) = \frac{3(1.99)^2 + 1}{(1.99 - 2)^2} = \frac{3(3.9601) + 1}{(0.01)^2} = \frac{11.8803 + 1}{0.0001} = \frac{12.8803}{0.0001} = 128803\]- For \(x = 1.999\): \[ r(1.999) = \frac{3(1.999)^2 + 1}{(1.999 - 2)^2} = \frac{3(3.996001) + 1}{(0.001)^2} = \frac{11.988003 + 1}{0.000001} = \frac{12.988003}{0.000001} = 12988003\]

Step 2: Evaluate Table 2

Substitute each value of \(x\) in Table 2 into the function \(r(x) = \frac{3x^2 + 1}{(x-2)^2}\) to find \(r(x)\).- For \(x = 2.5\): \[ r(2.5) = \frac{3(2.5)^2 + 1}{(2.5 - 2)^2} = \frac{3(6.25) + 1}{(0.5)^2} = \frac{18.75 + 1}{0.25} = \frac{19.75}{0.25} = 79\]- For \(x = 2.1\): \[ r(2.1) = \frac{3(2.1)^2 + 1}{(2.1 - 2)^2} = \frac{3(4.41) + 1}{(0.1)^2} = \frac{13.23 + 1}{0.01} = \frac{14.23}{0.01} = 1423\]- For \(x = 2.01\): \[ r(2.01) = \frac{3(2.01)^2 + 1}{(2.01 - 2)^2} = \frac{3(4.0401) + 1}{(0.01)^2} = \frac{12.1203 + 1}{0.0001} = \frac{13.1203}{0.0001} = 131203\]- For \(x = 2.001\): \[ r(2.001) = \frac{3(2.001)^2 + 1}{(2.001 - 2)^2} = \frac{3(4.004001) + 1}{(0.001)^2} = \frac{12.012003 + 1}{0.000001} = \frac{13.012003}{0.000001} = 13012003\]

Step 3: Behavioral Analysis Near Vertical Asymptote

From Tables 1 and 2, as \(x\) approaches 2 from both sides, \(r(x)\) tends to increase without bound, suggesting the existence of a vertical asymptote at \(x = 2\).

Step 4: Evaluate Table 3 and 4 (Horizontal Asymptote)

To find the horizontal asymptote, evaluate the behavior of \(r(x)\) as \(x\) approaches infinity and negative infinity, by using very large values in Tables 3 and 4.- For positive values (Table 3): As \(x = 10\), \(x = 50\), etc., compute: - For example, \(x = 10\): \[ r(10) = \frac{3(10)^2 + 1}{(10 - 2)^2} = \frac{301}{64} \approx 4.703125\] - The calculation for large \(x\) tends to approach 3, suggesting a horizontal asymptote of \(y = 3\).

Step 5: Conclusion

Based on Tables 3 and 4, it can be inferred that as \(x\) approaches positive or negative infinity, \(r(x)\) approaches \(3\), which confirms a horizontal asymptote at \(y = 3\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur in rational functions at values of \(x\) where the function becomes undefined or where the denominator equals zero. The equation we are considering is \( r(x) = \frac{3x^2 + 1}{(x-2)^2} \).
A vertical asymptote is at \(x = 2\) because plugging \(x = 2\) into the denominator \((x-2)^2\) results in zero, making the function undefined.
Observing Tables 1 and 2, as \(x\) approaches 2 from either side, the value of \(r(x)\) increases dramatically, suggesting the graph of the function shoots upwards toward infinity. This typical behavior indicates a vertical asymptote.

• Vertical asymptotes signify that the function grows unboundedly near \(x = 2\).
• They provide boundaries on graph, often making the function appear as if it curls away sharply.

When approaching vertical asymptotes, the approach can either be from positive or negative infinity, depending on the direction from which \(x\) approaches the asymptote.

Horizontal Asymptotes

Horizontal asymptotes show the behavior of a rational function as \(x\) tends to positive or negative infinity. For the function \( r(x) = \frac{3x^2 + 1}{(x-2)^2} \), identifying the horizontal asymptote helps us understand the function's end behavior.
Looking at Tables 3 and 4, where \(x\) takes on very large positive and negative values, we observe that \(r(x)\) approaches a value of 3.

• This tells us the horizontal asymptote is \(y = 3\).
• Horizontal asymptotes suggest outputs stabilize at a specific value as inputs become very large or very small.

In practical terms, as the variable \(x\) becomes extremely large in the positive or negative direction, the influence of the terms \(3x^2\) and \((x-2)^2\) cancel each other enough to let other terms dictate the approach, usually simple constants or dominant numerators and fewer dominance from denominators.

Behavior Near Asymptotes

Understanding how a function behaves near its asymptotes gives insight into its graph. Near vertical asymptotes, a function's output shoots towards infinity because the denominator approaches zero.
When \(x\) is near 2 in \( r(x) = \frac{3x^2 + 1}{(x-2)^2} \), the value of \(r(x)\) becomes exceedingly large on both sides, indicating sharp vertical growth. This means:

• Approaching from the left or right towards \(x = 2\) will result in extreme output values \((r(x) \to \infty)\).
• Graphically it appears as a significant spike or dip.

As \(x\) moves away from the vertical asymptote towards positive or negative infinity, the function settles around the horizontal asymptote \(y = 3\).
This steady leveling off happens gradually, with \(r(x)\) balancing towards 3 and reflects stabilization post away from extreme \(x\) values. This smoothing of the graph presents clear continuity as it contrasts sharply with near-asymptotal points.