Graphics calculators and rational functions

We can use technology such as graphing calculators to draw rational functions and identify certain properties of the function. This might include where the horizontal or vertical asymptotes lie, points of intersection, turning points and the domain and range of the function.

Exploration

Remember that a rational function is one that has the form

$f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$f(x)=p(x)q(x)​,

where $p\left(x\right)$p(x) and $q\left(x\right)$q(x) are polynomials and $q\left(x\right)\ne0$q(x)≠0.

Consider the rational function given by the equation $f\left(x\right)=\frac{x^2+2x-1}{\left(x+3\right)\left(x+5\right)}$f(x)=x2+2x−1(x+3)(x+5)​.

To draw the function using your graphing calculator, first locate the graphing feature of the calculator and enter the equation. From there, draw and view the graph as below.

The graph of $y=f\left(x\right)$y=f(x) and its asymptotes.

 

Immediately we can observe two vertical asymptotes with equations $x=-5$x=−5 and $x=-3$x=−3 and a horizontal asymptote at $y=1$y=1.

We can also identify the existence of a maximum and minimum turning point as indicated below.

Maximum and minimum turning points of the graph.

 

A graphing calculator can be used to approximate each turning point. The maximum and minimum turning points are $\left(-3.55,-5.65\right)$(−3.55,−5.65) and $\left(-1.79,-0.35\right)$(−1.79,−0.35) respectively.

Similarly we can identify points of intersection such as the $x$x-intercepts and $y$y-intercept. This function appears to pass through the origin, however on close inspection and using the graphing calculator, we can determine the $x$x-intercepts as $\left(-2.41,0\right),\left(0.41,0\right)$(−2.41,0),(0.41,0) and the $y$y-intercept as $\left(0,0.07\right)$(0,0.07) to two decimal places. Alternatively we can find the exact coordinates by substituting $x=0$x=0 and $y=0$y=0.

$x$x-intercepts marked above

 

$y$y-intercept marked above

 

Since the value of the function is undefined at $x=-5$x=−5 and $x=-3$x=−3 (the location of each vertical asymptote) we can state the domain of the function as $\left(-\infty,-5\right)\cup\left(-5,-3\right)\cup\left(-3,\infty\right)$(−∞,−5)∪(−5,−3)∪(−3,∞) which translates to all real numbers except for $x=\left\{-5,-3\right\}$x={−5,−3}.

Even though there's a horizontal asymptote, notice that the range is all real values, or $\left(-\infty,\infty\right)$(−∞,∞), since the function passes through the dotted line $y=1$y=1.

In general we can identify the points of intersection between any two functions, and not necessarily just the vertical and horizontal axes.

Consider the function $g\left(x\right)=x+5$g(x)=x+5 which we've drawn below.

The solution to the equation $\frac{x^2+2x-1}{\left(x+3\right)\left(x+5\right)}=x+5$x2+2x−1(x+3)(x+5)​=x+5 can be found by determining the $x$x-value of the point of intersection of the two graphs. In this case, the solution is $x=-2.77$x=−2.77 to two decimal places.

The graph of $f\left(x\right)=\frac{x^2+2x-1}{\left(x+3\right)\left(x+5\right)}$f(x)=x2+2x−1(x+3)(x+5)​ and $g\left(x\right)=x+5$g(x)=x+5.

 

Practice questions

Question 1

Question 2

Question 3