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After the numerator is divided by the denominator, $$f(x)=\frac{x^{5}+x^{4}+x^{2}+1}{x^{4}+1} \quad \text { becomes } \quad f(x)=x+1+\frac{x^{2}-x}{x^{4}+1}$$ (a) What is the oblique asymptote of the graph of the function? (b) Where does the graph of the function intersect its asymptote? (c) As \(x \rightarrow \infty,\) does the graph of the function approach its asymptote from above or below?

Short Answer

Expert verified
The oblique asymptote is \(y = x + 1\). Intersections are at (0,1) and (1,2). The graph approaches the asymptote from above for positive \(x\) and from below for negative \(x\).

Step by step solution

01

Identify the Oblique Asymptote

The given function is \(f(x)=\frac{x^{5}+x^{4}+x^{2}+1}{x^{4}+1}\). After simplifying the numerator divided by the denominator, it becomes \(f(x)=x+1+\frac{x^{2}-x}{x^{4}+1}\). In this form, the term \(x+1\) represents the oblique asymptote. Therefore, the oblique asymptote of the graph of the function is \(y = x + 1\).
02

Find the Intersection with the Asymptote

Set the function equal to its oblique asymptote to find the intersection points. This means we solve for \(f(x) = x + 1\). From the simplified form \(f(x) = x + 1 + \frac{x^{2} - x}{x^{4} + 1}\), set the remaining term \(\frac{x^{2} - x}{x^{4} + 1} = 0\). Solving for \(x\) we get: \begin{align*} x^{2} - x &= 0 \ \text{factor out x} \ x(x-1) &= 0 \ \text{solutions are} \ x &= 0 \ x &= 1 \ \text{Plug back into the asymptote} \ y &= 0 + 1 = 1 \ y &= 1 + 1 = 2 \ \text{Thus, the intersections are at (0, 1) and (1, 2).} \end{align*}
03

Analyze the Asymptotic Behavior

As \(x\) approaches infinity, the term \(\frac{x^{2} - x}{x^{4} + 1}\) in the simplified function \(f(x) = x+1 + \frac{x^{2} - x}{x^{4} + 1}\) becomes very small because the denominator grows faster than the numerator. Therefore, as \(x \rightarrow \infty\), \(\frac{x^{2} - x}{x^{4} + 1} \rightarrow 0\) and the function \(f(x)\) will approach the line \(y = x + 1\). To determine whether the graph approaches the asymptote from above or below, analyze the sign of \(\frac{x^{2} - x}{x^{4} + 1}\): - For large positive \(x\), as \(x^{2} - x\) is positive and the whole fraction is positive, so it's slightly above the asymptote. - For large negative \(x\), as \(x^{2} - x\) turns negative, the whole fraction must be negative, indicating the graph is slightly below the asymptote. \ Therefore, the graph approaches from above for positive \(x\)-values and from below for negative \(x\)-values.

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Key Concepts

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

asymptotic behavior
The concept of asymptotic behavior helps us understand how a function behaves as the variable approaches infinity. When we look at the fraction \(\frac{x^{2}-x}{x^{4}+1}\) in our given function, we notice something interesting. As \(x \rightarrow \infty\), the term \(\frac{x^{2}-x}{x^{4}+1}\) becomes very small. This is because the denominator \(x^{4}+1\) grows much faster than the numerator \(x^{2}-x\).

Thus, for very large values of \(x\), this entire fraction approaches zero. When a function has such a fractional component that diminishes as \(x\) grows larger, it helps in identifying the behavior near its asymptote.
The main takeaway: as \(x\) approaches infinity, \(f(x)\) gets closer and closer to \(x+1\). This is what we call asymptotic behavior. Understanding this simplifies predicting how the function behaves at extreme values, making graphing easier.
function intersection
Finding where a function intersects its asymptote can tell us more about its characteristics. We set our function equal to its oblique asymptote to find these points of intersection. For our function \(f(x) = x+1 + \frac{x^{2}-x}{x^{4}+1}\), we equate it to the asymptote \(x+1\).
This boils down to solving:
  • \(\frac{x^{2}-x}{x^{4}+1} = 0\)


Since fractions are zero only when their numerator is zero, we focus on \(x^2 - x = 0\).

By solving it, we get two \(x\) values:
  • \(x=0\)
  • \(x=1\)

Substituting these back into the function shows that:
  • At \(x=0\), \(y = 1\); so the point is (0, 1)
  • At \(x=1\), \(y = 2\); so the point is (1, 2)

Remember, these intersection points (0, 1) and (1, 2) show where the graph meets its asymptote.
end behavior of functions
End behavior tells us how the function acts as \(x\) goes to positive or negative infinity. For our function \(f(x) = x+1 + \frac{x^{2}-x}{x^{4}+1}\), as \(x\) becomes very large or very small, the term \(\frac{x^{2}-x}{x^{4}+1}\) becomes insignificant due to its denominator growing faster. Therefore, \(f(x)\) approaches the line \(y=x+1\).
To see if the function approaches from above or below, we examine the sign of \(\frac{x^{2}-x}{x^{4}+1}\):
  • For large positive \(x\): \(x^2 - x\) is positive, and hence the fraction is positive. This means \(f(x)\) is slightly above the asymptote.
  • For large negative \(x\): \(x^2 - x\) becomes negative, making the fraction negative. This means \(f(x)\) is slightly below the asymptote.

Therefore, as \(x \rightarrow \infty\), the graph approaches from above.
As \(x \rightarrow - \infty\), the graph approaches from below.

Understanding this helps in sketching the function’s graph accurately.

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