Problem 63 - Precalculus Mathematics for Calculus

Question

Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$s(x)=\frac{x^{2}-2 x+1}{x^{3}-3 x^{2}}$$

Step-by-Step Solution

Answer

x-intercept at (1,0); holes at x=0; vertical asymptote at x=3; horizontal asymptote at y=0; domain: x ≠ 0, 3; range: y ≠ 0.

Step 1: Find the x-intercepts

The x-intercepts occur where the numerator of the rational function equals zero. For the function \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), set the numerator equal to zero and solve for \( x \). \[ x^2 - 2x + 1 = 0 \] \( x^2 - 2x + 1 \) can be factored as \( (x-1)^2 = 0 \). So, the x-intercept is at \( x = 1 \).

Step 2: Find the y-intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \[ s(0) = \frac{0^2 - 2(0) + 1}{0^3 - 3(0)^2} = \frac{1}{0} \] Since division by zero is undefined, there is no y-intercept.

Step 3: Find the vertical asymptotes

Vertical asymptotes occur where the denominator equals zero but the numerator does not equal zero at the same points. Set the denominator of \( s(x) \) to zero and solve: \[ x^3 - 3x^2 = x^2(x - 3) = 0 \] Thus, \( x^2 = 0 \) gives \( x = 0 \) and \( x - 3 = 0 \) gives \( x = 3 \). At \( x = 0 \), the numerator is also zero, so it's not a vertical asymptote (hole instead). Therefore, the vertical asymptote is at \( x = 3 \).

Step 4: Find the horizontal asymptote

To find horizontal asymptotes, compare the degrees of the numerator and denominator. The numerator has a degree of 2, and the denominator has a degree of 3. Since the numerator's degree is less than the denominator's degree, the horizontal asymptote is \( y = 0 \).

Step 5: Determine any holes

Holes occur where factors cancel between the numerator and denominator. From step 1 and 3, the factor \( x^2 \) is common in both numerator and denominator at \( x = 0 \). Therefore, there is a hole at \( x = 1 \). However, there was a mistake here. The zero of the numerator corresponds exactly with an x-intercept, not a hole. Correctly, there is no hole further apart from the implicit behavior at x=0, which was misrepresented. Correction in concept, hence, x=0 is resolved as a hole due to factor \(x-0\) repeating in numerator & denominator outline.

Step 6: Sketch the graph and state domain and range

Using the information gathered: - x-intercept at \( (1, 0) \) - Vertical asymptote at \( x = 3 \) - Horizontal asymptote at \( y = 0 \) - Hole at \( x = 0 \). The domain is all real numbers except \( x = 0 \) and \( x = 3 \). The range is all real numbers except \( y = 0 \), with slight retrospective finding of horizontal readiness shift in graph so range narrow become emergent to reflect realistic curve estimates satify expositional intention.

Key Concepts

x-intercepts

When studying rational functions, x-intercepts are the points where the graph of the function crosses the x-axis. These are the values of \( x \) for which the function equals zero. To find the x-intercepts, set the numerator of the rational function to zero, since for fractions, if the numerator is zero, the whole expression becomes zero, assuming the denominator is not zero at that point.

For the function \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), to find the x-intercept, solve \( x^2 - 2x + 1 = 0 \).

This equation factors as \( (x - 1)^2 = 0 \), giving us the x-intercept at \( x = 1 \). This means the graph will touch the x-axis at this point.

Remember that when a factor is raised to an even power, like \((x - 1)^2 \), it means the graph will only "touch" the x-axis at that point, rather than crossing it.

vertical asymptotes

Vertical asymptotes are lines where the graph of a function approaches but does not touch or cross. They occur at the values of \( x \) that make the denominator zero, but not the numerator.

For the function \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), set the denominator to zero to find potential vertical asymptotes: \( x^3 - 3x^2 = x^2(x - 3) = 0 \).

This calculation gives \( x = 0 \) and \( x = 3 \). However, since \( x = 0 \) also makes the numerator zero, there is not a vertical asymptote at \( x = 0 \); instead, it is a hole.

Therefore, there is a vertical asymptote at \( x = 3 \), as the function will approach infinity as \( x \) nears 3 from either side.

horizontal asymptotes

Horizontal asymptotes represent the behavior of a graph as \( x \) approaches positive or negative infinity. They are determined by comparing the degrees of the numerator and the denominator in a rational function.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). For \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), the degree of the numerator (2) is less than the degree of the denominator (3).

Hence, the horizontal asymptote is \( y = 0 \). This means that as \( x \) gets very large or very small, the function value approaches zero. This helps in understanding the end behavior of the graph.

domain and range

The domain and range of a rational function describe the set of possible input values (x-values) and output values (y-values) of the function.

The domain includes all real numbers except where the denominator is zero, as division by zero is undefined. For the function \( s(x) = \frac{x^2 - 2x + 1}{x^3 - 3x^2} \), the domain is all real numbers except at \( x = 0 \) (a hole) and \( x = 3 \) (a vertical asymptote).

The range of a function is a bit trickier, especially with rational functions. The horizontal asymptote gives us an initial idea. For \( s(x) \), since \( y = 0 \) is a horizontal asymptote, \( y = 0 \) is not in the range because the function approaches but never reaches this value. Therefore, the range is all real numbers except \( y = 0 \).

Considering all this helps outline the extent to which the function's graph stretches across the coordinate plane.