5.05 Graphical and analytical solutions to rational equations

Graphing 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 the function passes through the dotted line $y=1$y=1. So the range is all values of $y$y except for those between the maximum value of the middle branch and the minimum value of the right branch of the function. That is, $\left(-\infty,-5.65\right)\cup\left(-0.35,\infty\right)$(−∞,−5.65)∪(−0.35,∞)

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

 

Solve rational functions analytically

A variety of equations arise from time to time that involves the sums and differences of rational expressions. The solution of these almost always begins by multiplying both sides of the equation by well-chosen factors that will 'clear' the denominators.

Worked examples

Question 4

Solve $\frac{x}{x-5}+\frac{10}{x+5}=\frac{4}{x^2-25}$xx−5​+10x+5​=4x2−25​

Observe that $x^2-25=\left(x-5\right)\left(x+5\right)$x2−25=(x−5)(x+5), and so the strategy becomes clear. We need to multiply both sides of the equation by $x^2-25$x2−25 to clear the denominators.

$\frac{x}{x-5}+\frac{10}{x+5}$xx−5​+10x+5​	$=$=	$\frac{4}{x^2-25}$4x2−25​
$\left(x-5\right)\left(x+5\right)\frac{x}{\left(x-5\right)}+\left(x-5\right)\left(x+5\right)\frac{10}{\left(x+5\right)}$(x−5)(x+5)x(x−5)​+(x−5)(x+5)10(x+5)​	$=$=	$\left(x-5\right)\left(x+5\right)\frac{4}{x^2-25}$(x−5)(x+5)4x2−25​
$x\left(x+5\right)+10\left(x-5\right)$x(x+5)+10(x−5)	$=$=	$4$4
$x^2+15x-54$x2+15x−54	$=$=	$0$0
$\left(x+18\right)\left(x-3\right)$(x+18)(x−3)	$=$=	$0$0
$\therefore$∴     $x$x	$=$=	$3,-18$3,−18

 

question 5

Solve $\frac{48}{x^2-8x}-\frac{80}{x^2-64}=0$48x2−8x​−80x2−64​=0

Here we need to notice that the factor $\left(x-8\right)$(x−8) is common to both denominator expressions. Thus multiplying both sides of the equation by $x\left(x-8\right)\left(x+8\right)$x(x−8)(x+8) will clear the denominators. 

$\frac{48}{x^2-8x}-\frac{80}{x^2-64}$48x2−8x​−80x2−64​	$=$=	$0$0
$\frac{48}{x\left(x-8\right)}-\frac{80}{\left(x-8\right)\left(x+8\right)}$48x(x−8)​−80(x−8)(x+8)​	$=$=	$0$0
$x\left(x-8\right)\left(x+8\right)\frac{48}{x\left(x-8\right)}-x\left(x-8\right)\left(x+8\right)\frac{80}{\left(x-8\right)\left(x+8\right)}$x(x−8)(x+8)48x(x−8)​−x(x−8)(x+8)80(x−8)(x+8)​	$=$=	$0$0
$48\left(x+8\right)-80x$48(x+8)−80x	$=$=	$0$0
$32x$32x	$=$=	$384$384
$\therefore$∴    $x$x	$=$=	$12$12

  

question 6

Find all real solutions to the equation  $x^{-4}+7x^{-2}-144=0$x−4+7x−2−144=0. Are there any complex solutions?

To find the real solutions we could proceed as follows:

Re-write the equation as $\frac{1}{x^4}+\frac{7}{x^2}-144=0$1x4​+7x2​−144=0, and then multiply both sides by $x^4$x4. This would lead to the new equation $1+7x^2-144x^4=0$1+7x2−144x4=0. 

Then by setting $u=x^2$u=x2, and rearranging, we establish the quadratic equation given by $144u^2-7u-1=0$144u2−7u−1=0. We solve this as follows:

$144u^2-7u-1$144u2−7u−1	$=$=	$0$0
$\left(16u+1\right)\left(9u-1\right)$(16u+1)(9u−1)	$=$=	$0$0
$\therefore$∴     $u$u	$=$=	$-\frac{1}{16},\frac{1}{9}$−116​,19​

For $u=x^2=-\frac{1}{16}$u=x2=−116​, there is no real solution because, for all $x$x, $x^2\ge0$x2≥0. 

For $u=x^2=\frac{1}{9}$u=x2=19​, we have that $x=\pm\frac{1}{3}$x=±13​.

NOTE:

A slightly simpler alternative approach to the problem is to immediately set $u=x^{-2}$u=x−2 so that the equation $x^{-4}+7x^{-2}-144=0$x−4+7x−2−144=0 changes to $u^2+7u-144=0$u2+7u−144=0.

Factoring, we have $\left(u+16\right)\left(u-9\right)=0$(u+16)(u−9)=0 and therefore $u=-16,9$u=−16,9.

Setting $u=x^{-2}=-16$u=x−2=−16, and $u=x^{-2}=9$u=x−2=9, we again arrive at the same conclusion that the real solutions are given by $x=\pm\frac{1}{3}$x=±13​. 

If we admit complex solutions, then we can solve the equation $x^2=-\frac{1}{16}$x2=−116​ as follows:

$x^2$x2	$=$=	$-\frac{1}{16}$−116​
$16x^2+1$16x2+1	$=$=	$0$0
$16x^2-i^2$16x2−i2	$=$=	$0$0
$\left(4x+i\right)\left(4x-i\right)$(4x+i)(4x−i)	$=$=	$0$0
$\therefore$∴     $x$x	$=$=	$\pm\frac{i}{4}$±i4​

 

Practice questions

Question 7

Question 8

Question 9