Problem 50 - A Graphical Approach to College Algebra

Question

Sketch a graph of each rational function. Your graph should include all asymptotes. Do not use a calculator. $$f(x)=\frac{(3-x)^{2}}{(1-x)(4+x)}$$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x=1\) and \(x=-4\); horizontal asymptote at \(y=-1\); x-intercept at \(x=3\); y-intercept at \((0,\frac{9}{4})\).

Step 1: Identify Vertical Asymptotes

A rational function has vertical asymptotes where its denominator becomes zero provided the numerator is not also zero at these points. Set the denominator equal to zero: \((1-x)(4+x) = 0\). Solve these to find the vertical asymptotes:- For \(1-x=0\), solvinng gives \(x=1\).- For \(4+x=0\), solving gives \(x=-4\).Thus, the vertical asymptotes are at \(x=1\) and \(x=-4\).

Step 2: Identify Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the polynomial in the numerator and the polynomial in the denominator. Here, the numerator \((3-x)^2\) is a quadratic polynomial and the denominator \((1-x)(4+x)\) can be expanded to \(-(x^2+3x-4)\), also a quadratic. Since the degrees of the numerator and denominator are equal, the horizontal asymptote is the leading coefficient of the numerator divided by the leading coefficient of the denominator. Thus, the horizontal asymptote is \(y=\frac{1}{-1}=-1\).

Step 3: Identify the X-Intercepts

The x-intercepts occur where the numerator is zero, provided the denominator is not zero. Set the numerator equal to zero:\((3-x)^2=0\).Solving gives \(3-x=0\) or \(x=3\).Hence, there is an x-intercept at \(x=3\).

Step 4: Identify the Y-Intercept

The y-intercept occurs where x=0. Substitute x=0 into the function to find the y-value:\[f(0)=\frac{(3-0)^2}{(1-0)(4+0)}=\frac{9}{4}\].Thus, the y-intercept is at \((0,\frac{9}{4})\).

Step 5: Confirm Bounded Behavior Around Asymptotes

Analyze how the function behaves around the vertical asymptotes by testing intervals. For this function, determine the sign changes around \(x=1\) and \(x=-4\). Test intervals are \((-\infty, -4)\), \((-4, 1)\), and \((1, \infty)\). Substitute test points such as \(x=-5, 0, 2\) into the numerator and denominator to find the sign of \(f(x)\). This confirms the behavior of the function near the asymptotes and helps sketch the graph. Note there are no sign changes at \(x=3\) because it is a double root, confirming the x-intercept behavior.

Key Concepts

Vertical Asymptotes

A rational function may have vertical asymptotes, which are vertical lines where the function becomes undefined. These occur where the denominator is zero but the numerator isn't, indicating points of infinite behavior. In our exercise, setting the denominator \((1-x)(4+x) = 0\) helps locate these vertical asymptotes. Doing this, we solve two simple equations:

• For \(1-x=0\), solve to get \(x=1\).
• For \(4+x=0\), solve to get \(x=-4\).

Thus, the rational function\(f(x)=\frac{(3-x)^{2}}{(1-x)(4+x)}\) has vertical asymptotes at \(x = 1\) and \(x = -4\). Vertical asymptotes show abrupt changes in the graph, splitting the pressure points where the function's value climbs or dives to infinity, providing clear indicators during graph sketching.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of rational functions as \(x\) approaches infinity. They show where the function flattens out. Identifying them involves comparing the degrees of the polynomials in the numerator and the denominator.
The rule is simple:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y=0\).
• If the degrees are equal, as in our exercise, the horizontal asymptote is found by dividing the leading coefficients of the numerator by that of the denominator.
• If the degree of the numerator exceeds that of the denominator, there's no horizontal asymptote, though there may be a slant asymptote.

In this instance, both degrees are the same. So, we compare the leading coefficients giving us the horizontal asymptote at\(y=-1\). This line acts as an invisible boundary that the graph approaches but never crosses for extremely large positive or negative values of \(x\).

X-Intercepts

X-intercepts are points where the graph crosses the x-axis, meaning the function value is zero there. To find these, set the numerator of the rational function to zero since the denominator cannot be zero at these points as well, indicating exact zero crossing points.
For the function in our exercise, the numerator is given by \((3-x)^2 = 0\). Solving this, we simplify to \(3-x=0\), leading us to find that \(x=3\).
Thus, the x-intercept is at the point \((3, 0)\). As this implies the graph of \(f(x)\) touches or crosses the x-axis here, it's crucial for understanding the function's root structure and visual layout. However, note that in this case, this intercept corresponds to a double root, meaning no sign change.

Y-Intercept

The y-intercept is where the function crosses the y-axis, which happens when \(x=0\). To find it, simply substitute \(x=0\) into the function and calculate the resultant y-value. This gives a clear starting point for graphing.
Applying this to our rational function:\[f(0)=\frac{(3-0)^2}{(1-0)(4+0)}=\frac{9}{4}\].Thus, the y-intercept is \((0, \frac{9}{4})\). This point helps define the vertical positioning of the graph, especially close to the origin. It acts as a useful anchor or starting point when plotting the graph of a rational function.