Rational Functions

$$f(x)=\frac{x^{2}-5x-24}{x^{2}-8x-33}$$

Solution:

In this case, we have \(P(x)=x^{2}-5x-24\) and \(Q(x)=x^{2}-8x-33\).

To see if we can simplify this rational function, we can factor \(x^{2}-5x-24\) as \((x+3)(x-8)\).

Similarly \(Q(x)\) can be rewritten as \((x+3)(x-11)\).

Putting that altogether gives us the following:

$$f(x)=\frac{(x+3)(x-8)}{(x+3)(x-11)}$$

Notice that the numerator and the denominator of this rational expression have as common factor \((x+3)\). Therefore, these factors cancel each other out, leaving us with only \((x-8)\)in the numerator and \((x-11)\)in the denominator.

$$f(x)=\left(\frac{x-8}{x-11}\right)$$

Simplify the following rational function:

$$\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}$$

Solution:

Let the two be functions as \(p(x)=x^{2}+10x+24\) and \(q(x)=x^{3}-x^{2}-20x\)

We will simplify both of them individually and then take the ratio of them at the end.

Factorising \(p(x)\):

$$\begin{split} p(x) &=x^2 +10x+24 \\ &=x^2 +6x+4x+24 \\ &=x(x+6)+4(x+6) \\ &=(x+4)(x+6) \end{split}$$

All we did is splitting the middle term and then taking out the common factor.

Now, factorising \(q(r)\):

$$\begin{split} q(x) &=x^3 -x^2-20x \\ &=x(x^2 - x -20) \\ &=x(x^2 -5x+4x-20) \\ &=x(x-5)(x+4) \end{split}$$

Here we took out \(x\) as a common factor and then split the middle term.

Therefore, we can evaluate \(\frac{x^2 +10x+24}{x^3 -x^2 -20}\):

$$\begin{split} \frac{x^2 +10x+24}{x^3 -x^2 -20} &=\frac{(x+4)(x+6)}{x(x-5)(x+4)} \\ &=\frac{x+6}{x(x-5)} \end{split}$$

1. We start with a rational function:

$$f(x)=\frac{x-8}{x-11}$$

2. Next we need to replace the \(f(x)\) with \(y\):

$$y=\frac{x-8}{x-11}$$

3. Every \(x\) and \(y\) need to be swapped around:

$$x=\frac{y-8}{y-11}$$

4. We now solve this equation for \(y\) in terms of \(x\). This will take a few steps, as it is not something we can complete in one go. The first thing to do is multiply both sides by \((y-11)\):

$$y-8=x(y-11)$$

Then we distribute \(x\) on the right-hand side:

$$y-8=xy-11x$$

Then, we subtract \(xy-8\) from both sides:

$$y(1-x)=8-11x$$

Finally, we can divide both sides by \(1-x\):

$$y=\frac{8-11x}{1-x}$$

5. The last thing we need to do is replace \(y\) with \(f^{-1}(x)\) which indicates the fact that it is the inverse of the original rational function \(f(x)\).

$$f^{-1}(x)=\frac{8-11x}{1-x}$$

To check whether the obtained answer is correct or not, we can take a value of \(x\) and evaluate it at \(f(x)\):

$$f(0)=\frac{0-8}{0-11}=\frac{8}{11}$$

Now we know that \(f(0)=8/11\), so we need to check if we get \(f^{-1}(8/11)=0\) from our inverse equation:

$$f^{-1}\left(\frac{8}{11}\right)=\frac{8-11 \left( \frac{8}{11} \right)}{1-\frac{8}{11}}=0$$

Hence our answer is indeed correct.

Draw the graph the following function.

$$f(x)=\frac{2x^{2}+x+1}{x+1}$$

Solution:

1. Because \(Q(x)=x+1\) becomes \(0\) at \(x=-1\), we can tell that there will be a vertical asymptote at that x-value. Furthermore, since the degree of the numerator is greater than the degree of the denominator by exactly one. Therefore, there is an oblique asymptote, and consequently, there are no horizontal asymptotes.

To find the equation of the oblique asymptote, we need to do the polynomial division:

$$\frac{2x^{2}+x+1}{x+1}=2x-1+\frac{2}{x+1}$$

Since the equation of the asymptote must be of the form \(y=mx+b\), we can identify the asymptote equation to be \(2x-1\).

2. We draw these asymptotes on our paper. In this case, the vertical line is our vertical asymptote of \(x=-1\) and our slant line is our oblique asymptote of \(2x-1\):

Fig. 2 - A graph showing the asymptotes of a rational function - StudySmarter Originals

3. To find any intercepts we need to set \(x\) to \(0\) and then \(y\) to \(0\). First, we will set \(x\) to \(0\):

$$\frac{2(0)^{2}+0+1}{0+1}$$

which simplifies down to \(\frac{0+0+1}{0+1}\). Therefore there is an intercept with the y axis at \(y = 1\). Now, setting \(y=0\), we obtain the equation

$$0=\frac{2(x)^{2}+x+1}{x+1}$$

where we can ignore the denominator, leaving us with \(2(x)^{2}+x+1=0\). The reason why we can eliminate the denominator is that if we multiply both sides with it, multiplying with \(0\) will vanish it anyways. This equation has no real solution. This implies that there is no point where the parabola of \(2(x)^{2}+x+1=0\) crosses the x-axis. Therefore, there are no x-intercepts. Make sure to plot these points on the graph.

Fig. 3 - The graph involving the asymptotes and the y intercept - StudySmarter Originals

4. We now plot several values for \(x\) against \(y\), substituting an \(x\) value into the equation and plotting the point on the graph:

Fig. 4 - The graph after plotting some more points - StudySmarter Originals

5. Finally, we can connect these points to draw our final graph:

Fig. 5 - The curve after joining all the points and asymptotes addressed - StudySmarter Originals

Notice that the graph is split up into two different pieces. This will happen with every rational function.

More examples on Graphing Rational Functions can be found here!