Rational Function Analysis Exercise 5
We're tasked with analyzing the graph of the following rational function:
$$ f(x) = \frac{5x-1}{x^2-3x+2} $$
To do this, we'll apply standard tools from calculus, working through each step methodically.
Domain
We begin by determining the domain of the function.
This function is defined for all real values of \( x \), except where the denominator is zero.
The denominator \( x^2 - 3x + 2 \) factors easily, and its roots are:
$$ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} $$
$$ x = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2} $$
$$ x = \begin{cases} \frac{2}{2} = 1 \\[0.5em] \frac{4}{2} = 2 \end{cases} $$
Therefore, the domain of \( f \) is:
$$ D_f = (-\infty, 1) \cup (1, 2) \cup (2, \infty) $$
The function is undefined at \( x = 1 \) and \( x = 2 \).
Intercepts
To find the y-intercept, we evaluate the function at \( x = 0 \):
$$ f(0) = \frac{5 \cdot 0 - 1}{0^2 - 3 \cdot 0 + 2} = \frac{-1}{2} $$
So the graph crosses the y-axis at the point (0, -1/2).
For the x-intercept, we solve \( f(x) = 0 \):
$$ \frac{5x - 1}{x^2 - 3x + 2} = 0 $$
A rational expression is zero when its numerator is zero (as long as the denominator isn't):
$$ 5x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{5} $$
Thus, the graph intersects the x-axis at (1/5, 0).
We can now mark these two key points on the coordinate plane:
Sign Analysis
Next, we analyze the sign of the function.
We'll study the signs of the numerator and the denominator separately:
- The numerator \( 5x - 1 \) is positive for \( x > \frac{1}{5} \), zero at \( x = \frac{1}{5} \), and negative for \( x < \frac{1}{5} \).
- The denominator \( x^2 - 3x + 2 \) is a quadratic opening upwards, with roots at \( x = 1 \) and \( x = 2 \). It's positive on \( (-\infty, 1) \cup (2, \infty) \), and negative on \( (1, 2) \).
Combining these results, the function is:
- Negative on \( (-\infty, \frac{1}{5}) \) and \( (1, 2) \)
- Positive on \( (\frac{1}{5}, 1) \) and \( (2, \infty) \)
We use this information to exclude the regions of the graph where the function doesn’t exist or takes on inadmissible signs (shaded in gray):
Asymptotes
Horizontal Asymptote
We analyze the behavior of the function as \( x \) tends to infinity:
$$ \lim_{x \to \infty} \frac{5x - 1}{x^2 - 3x + 2} = \frac{\infty}{\infty} $$
This is an indeterminate form, so we apply L’Hôpital’s Rule:
$$ \lim_{x \to \infty} \frac{d}{dx}[5x - 1] \Big/ \frac{d}{dx}[x^2 - 3x + 2] = \lim_{x \to \infty} \frac{5}{2x - 3} = 0^+ $$
As \( x \to \infty \), the function approaches 0 from above.
Now consider the behavior as \( x \to -\infty \):
$$ \lim_{x \to -\infty} \frac{5x - 1}{x^2 - 3x + 2} = \frac{-\infty}{\infty} $$
Again, we apply L’Hôpital’s Rule:
$$ \lim_{x \to -\infty} \frac{5}{2x - 3} = 0^- $$
As \( x \to -\infty \), the function approaches 0 from below.
Therefore, the function has a horizontal asymptote along the x-axis: \( y = 0 \).
