Rectangular hyperbolas have the standard general form given by $y=\frac{a}{x-h}+k$y=ax−h+k, where $a$a is called the dilation factor, and $h$h and $k$k are the horizontal and vertical translation constants respectively. for more information on how these transformation constants effect the graph, please refer to this entry.
For example the hyperbola given by $y=\frac{3}{x-7}$y=3x−7 has the dilation factor given by $a=3$a=3, and the translation constants $h=7$h=7 and $k=0$k=0.
As another less obvious example, we can show that the curve given by $\left(x-2\right)\left(y+4\right)=-5$(x−2)(y+4)=−5 is also an hyperbola:
$\left(x-2\right)\left(y+4\right)$(x−2)(y+4) | $=$= | $-5$−5 |
$\left(y+4\right)$(y+4) | $=$= | $\frac{-5}{\left(x-2\right)}$−5(x−2) |
$y$y | $=$= | $\frac{-5}{\left(x-2\right)}-4$−5(x−2)−4 |
Here, $a=-5$a=−5, $h=2$h=2 and $k=-4$k=−4, so that the centre of the hyperbola is located at the point $\left(2,-4\right)$(2,−4), the two mutually perpendicular asymptotes have equations $x=2$x=2 and $y=-4$y=−4, and the two arcs are in the top left and bottom right quadrants created by these asymptotes.
The hyperbola $y=\frac{x-5}{x+2}$y=x−5x+2 can be expressed in general form as follows:
$y$y | $=$= | $\frac{x-5}{x+2}$x−5x+2 |
$=$= | $\frac{\left(x+2\right)-7}{x+2}$(x+2)−7x+2 | |
$=$= | $\frac{x+2}{x+2}-\frac{7}{x+2}$x+2x+2−7x+2 | |
$=$= | $\frac{x+2}{x+2}-\frac{7}{x+2}$x+2x+2−7x+2 | |
$=$= | $y=\frac{-7}{x+2}+1$y=−7x+2+1 | |
Here, $a=-7$a=−7 and the asymptotes are given by the lines $x=-2$x=−2 and $y=1$y=1.
Our final example is the curve given by $y=\frac{9}{2x-3}+5$y=92x−3+5. We can express this as:
$y$y | $=$= | $\frac{9}{2x-3}+5$92x−3+5 |
$=$= | $\frac{9}{2\left(x-\frac{3}{2}\right)}+5$92(x−32)+5 | |
$=$= | $\frac{\frac{9}{2}}{\left(x-\frac{3}{2}\right)}+5$92(x−32)+5 | |
$=$= | $y=\frac{4.5}{\left(x-1.5\right)}+5$y=4.5(x−1.5)+5 | |
Thus $a=4.5$a=4.5, $h=1.5$h=1.5 and $k=5$k=5.
Every rectangular hyperbola has two mutually perpendicular asymptotes given by $x=h$x=h and $y=k$y=k. Thus $x=h$x=h is the only point excluded from the domain and $y=k$y=k is the only point excluded from the range. We usually state this formally as, in the case of the domain, $x:x\in\Re,x\ne h$x:x∈ℜ,x≠h and in the case of the range, $y:y\in\Re,y\ne k$y:y∈ℜ,y≠k.
In other words, expressing a rectangular hyperbola in its general form reveals the natural exclusions from its domain and range.
Looking back at example $4$4, with $y=\frac{9}{2x-3}+5$y=92x−3+5, we know that the denominator of the fractional part cannot be $0$0. Thus solving $2x-3=0$2x−3=0, we see that $x=\frac{3}{2}=1.5$x=32=1.5 and so the domain is given by all real numbers not equal to $1.5$1.5.
The range is straightforward as well. If we consider $y=\frac{9}{2x-3}+5$y=92x−3+5 logically, we realise that y can never be $5$5, because that would force $\frac{9}{2x-3}$92x−3 to be $0$0, and no value of $x$x will make that occur because the numerator is always $9$9 (that is to say, a fraction that has the value $0$0 must have its numerator $0$0). So, the range is given by all real values of $y$y not equal to $0$0.
Formally, we can state, for the domain, $x:x\in\Re,x\ne1.5$x:x∈ℜ,x≠1.5 and for the range, $y:y\in\Re,y\ne5$y:y∈ℜ,y≠5.
In example $3$3, for the hyperbola $y=\frac{x-5}{x+2}$y=x−5x+2, we immediately see that $x=2$x=2 is excluded because no real fraction can have a zero denominator.
We might also argue that as $x$x becomes larger and larger, the fraction becomes closer to $1$1. For example, with $x=1000$x=1000, $y=\frac{995}{1002}$y=9951002 and with $x=100,000$x=100,000, $y=\frac{99995}{100002}$y=99995100002. This means that the range excludes the value $y=-1$y=−1.
Of course, even though the fractions are getting closer to $1$1, can we be sure about this? It could be that the fractions are getting closer to some number very near $1$1, but not exactly $1$1.
Perhaps a more clearer mathematical argument would be to change $y=\frac{x-5}{x+2}$y=x−5x+2 into $y=\frac{-7}{x+2}+1$y=−7x+2+1 (see above) and then argue that as $x$x gets larger and larger, the fraction $\frac{-7}{x+2}$−7x+2 gets smaller and smaller, tending toward $0$0, so that the limiting value is indeed $1$1.
Thus the domain is given by $x:x\in\Re,x\ne2$x:x∈ℜ,x≠2 and for the range, $y:y\in\Re,y\ne1$y:y∈ℜ,y≠1.
In example $2$2 above we had the curve given by $\left(x-2\right)\left(y+4\right)=-5$(x−2)(y+4)=−5.
Logically, this expression is stating that the product of two things, namely $\left(x-2\right)$(x−2) and $\left(y+4\right)$(y+4) has the value $-5$−5. If either or both of those things become $0$0, then their product can never be $-5$−5. In fact, their product becomes zero.
Thus we cannot let either $x-2=0$x−2=0 or $y+4=0$y+4=0. Hence $x=2$x=2 and $y=-4$y=−4 are the exclusions. Formally, we can state that the domain is given by $x:x\in\Re,x\ne2$x:x∈ℜ,x≠2 and for the range, $y:y\in\Re,y\ne-4$y:y∈ℜ,y≠−4.
Perhaps the best way forward is to know both approaches. When we recognise the general form of an hyperbola, we can simply state the domain and range. When an hyperbola is not in general form, we can either manipulate the equation so that it is in general form, or look at the given expression logically and reason our way to the exclusions.
Note that the dilation factor of a hyperbola can be more obscure when the equation is not given in general form.
Consider the function $y=\frac{2}{x}$y=2x.
State the domain of the function.
All real values of $x$x such that $x>0$x>0.
$x=0$x=0 only.
All real values of $x$x.
All real values of $x$x, except $x=0$x=0.
All real values of $x$x such that $x>0$x>0.
$x=0$x=0 only.
All real values of $x$x.
All real values of $x$x, except $x=0$x=0.
State the equation of the vertical asymptote.
Rearrange the equation to express $x$x in terms of $y$y.
Hence state the range of the function.
$y=0$y=0 only.
All real values of $y$y, except $y=0$y=0.
All real values of $y$y.
All real values of $y$y such that $y>0$y>0.
$y=0$y=0 only.
All real values of $y$y, except $y=0$y=0.
All real values of $y$y.
All real values of $y$y such that $y>0$y>0.
Hence state the equation of the horizontal asymptote.
Consider the function $y=\frac{2}{x-4}+3$y=2x−4+3.
Fill in the gap to state the domain of the function.
domain$=$={$x$x$\in$∈$\mathbb{R}$ℝ; $x\ne\editable{}$x≠}
State the equation of the vertical asymptote.
As $x$x approaches $\infty$∞, what value does $y$y approach?
Hence state the equation of the horizontal asymptote.
State the range of the function.
range$=$={$y$y$\in$∈$\mathbb{R}$ℝ; $y\ne\editable{}$y≠}
Which of the following is the graph of the function?
Consider the function $y=\frac{x-2}{x-4}$y=x−2x−4.
Fill in the gap to create equivalent expressions.
$\frac{x-2}{x-4}=\frac{x-4+\editable{}}{x-4}$x−2x−4=x−4+x−4
Hence express $y=\frac{x-2}{x-4}$y=x−2x−4 in the form $y=\frac{m}{x-h}+k$y=mx−h+k, for some values $k$k and $h$h.
State the equation of the vertical asymptote.
As $x$x approaches $\infty$∞, what does $y$y approach?
Hence state the equation of the horizontal asymptote.
Which of the following is the graph of the function?
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
Apply graphical methods in solving problems