9.01 Transformations of 1/x

Graphing y = k/x

We can graph a function $\frac{k}{x}$kx​ by constructing a table of values having first specified a value for the parameter $k$k. The shape of the graph is displayed below and the effect of changing $k$k is to change the scale of the graph. These properties are illustrated in the following diagram where the graph of $y=\frac{1}{x}$y=1x​ is shown in blue,  $y=\frac{3}{x}$y=3x​ is shown in red and $y=\frac{5}{x}$y=5x​ is shown in green.

If we had to draw these graphs by hand, we could construct tables of values like the following. We have restricted $x$x to values between $-5$−5 and $5$5.

You should check whether the graphs above really do match the corresponding tables.

 

$x$x	$-5$−5	$-4$−4	$-3$−3	$-2$−2	$-1$−1	$\frac{1}{2}$12​	$1$1	$2$2	$3$3	$4$4	$5$5
$\frac{1}{x}$1x​	$-\frac{1}{5}$−15​	$-\frac{1}{4}$−14​	$-\frac{1}{3}$−13​	$-\frac{1}{2}$−12​	$-1$−1	$2$2	$1$1	$\frac{1}{2}$12​	$\frac{1}{3}$13​	$\frac{1}{4}$14​	$\frac{1}{5}$15​
$\frac{3}{x}$3x​	$-\frac{3}{5}$−35​	$-\frac{3}{4}$−34​	$-1$−1	$-\frac{3}{2}$−32​	$-3$−3	$6$6	$3$3	$\frac{3}{2}$32​	$1$1	$\frac{3}{4}$34​	$\frac{3}{5}$35​
$\frac{5}{x}$5x​	$-1$−1	$-\frac{5}{4}$−54​	$-\frac{5}{3}$−53​	$-\frac{5}{2}$−52​	$-5$−5	$10$10	$5$5	$\frac{5}{2}$52​	$\frac{5}{3}$53​	$\frac{5}{4}$54​	$1$1

Practice question

Question 1

 

 

Identifying the shape of y = k/x

Imagine a sequence of rectangles drawn so that their heights are decreasing while their widths are increasing. They have been constructed so that their areas are all the same.

We could write $hw=A$hw=A where $h$h and $w$w are variables and $A$A is a constant.

Equivalently, we could draw a graph in the $xy$xy-plane that shows the points whose product $xy$xy is a particular constant $k$k. We obtain the graph of a relation $xy=k$xy=k.

Try exploring this applet, where the top slider changes the area of the rectangle, and the bottom slider changes one dimension of the rectangle. The width changes automatically to ensure that the area of the rectangle stays constant.

While this applet only shows the graph in the first quadrant, you can observe how changing the value of $k$k changes the shape of $y=\frac{k}{x}$y=kx​ in a dynamic way. If we look at all the quadrants, we obtain the full graph of the relation $xy=2$xy=2, and you may be more familiar seeing it written as $y=\frac{2}{x}$y=2x​

 

This graph gives more than just the dimensions of rectangles with area $2$2. It also gives the points $xy=2$xy=2 when $x$x and $y$y can be negative, even though we do not normally allow negative lengths and widths of geometrical objects.

Key Characteristics of the Parent Function $f\left(x\right)=\frac{1}{x}$f(x)=1x​:

Characteristic	$f\left(x\right)=\frac{1}{x}$f(x)=1x​
Domain	Inequality form: $x\ne0$x≠0 Interval form: $\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞)
Range	Inequality form: $y\ne0$y≠0 Interval form: $\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞)
Extrema	None
$x$x-intercept	None
$y$y-intercept	None
Increasing/decreasing	Decreasing over its domain, $x\ne0$x≠0
End behavior	As $x\to-\infty$x→−∞, $y\to0$y→0, and as $x\to\infty$x→∞, $y\to0$y→0
Asymptotes	Vertical: $x=0$x=0 and Horizontal: $y=0$y=0

 

Practice questions

Question 2

Question 3

 

Transformations of $y=\frac{k}{x}$y=kx​

Expressing in general form

Transformations of $y=\frac{k}{x}$y=kx​  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. 

Everything we have already learned about transformations still applies.

For example the graph of $g(x)=\frac{3}{x-7}$g(x)=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.

Applet

Use the applet to understand how the translation of the basic function works. Try positive and negative values of $a$a. 

Worked example

Question 4

Describe the transformation which have occur from $y=\frac{1}{x}$y=1x​ to $\left(x-2\right)\left(y+4\right)=-5$(x−2)(y+4)=−5.

Think: We first need to re-arrange the transformed function to the form $y=\frac{a}{x-h}+k$y=ax−h​+k.

Do:

$\left(x-2\right)\left(y+4\right)$(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 center of the graph is located at the point. It has been stretched vertically by a factor of $5$5, reflected in the $x$x-axis, translated right $3$3 and down $4$4. This will make the asymptotes $x=2$x=2 and $y=-4$y=−4, and the two arcs are in the top left and bottom right quadrants.

 

Domain and range

Every transformation of $y=\frac{k}{x}$y=kx​ 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 the equation in its general form reveals the natural exclusions from its domain and range.

 

Practice questions

Question 5

Question 6

 

Graphing transformations of $y=\frac{k}{x}$y=kx​

The graph of every transformation of $y=\frac{k}{x}$y=kx​ has two separate arcs and these arcs in their extremities move toward straight lines called asymptotes.

Every transformation of $y=\frac{k}{x}$y=kx​, just like every circle, has a center but, just like a circle, this center is not part of the curve itself. One important feature of every graph that represents a transformation of $y=\frac{k}{x}$y=kx​ is that if we rotate any one of the arcs by $180^{\circ}$180∘ about the center of the graph, that arc will fall directly onto the other arc. 

The center of the graph given by $y=\frac{a}{x-h}+k$y=ax−h​+k is the point $\left(h,k\right)$(h,k). It provides both the key to the asymptotes and to the efficient plotting of the curve.

The asymptotes of a transformation of $y=\frac{k}{x}$y=kx​ are given by the vertical line $x=h$x=h (this is the line passing through the point $\left(h,0\right)$(h,0) and parallel to the $y$y axis) and the horizontal line $y=k$y=k (this is the line passing through the point $\left(0,k\right)$(0,k) and parallel to the $x$x axis).

To ensure a  a good plot of the graph we should select a range $x$x values on either side of the asymptote $x=h$x=h.

 

Worked example

QUestion 7

One improvement we could make to our strategy is to choose $x$x values that, while not equally separated, give rise to integer $y$y values whenever possible. 

For example suppose we wish to plot the graph of $y=\frac{12}{x-3}+7$y=12x−3​+7. We note that the center is $\left(3,7\right)$(3,7), and the two mutually perpendicular asymptotes are the straight lines $x=3$x=3 and $y=7$y=7. 

As in the first example, we choose an equal number of $x$x values on either side of the line $x=3$x=3. However this time we will choose the $x$x values so that the quantity $\left(x-3\right)$(x−3) is a factor of $12$12. This ensures that the $y$y values will be integers.

For example we might choose $x=4$x=4 because $\frac{12}{4-3}+7$124−3​+7 is the integer $19$19. Similarly the five values $x=5$x=5, $x=6$x=6, $x=7$x=7, $x=9$x=9, $x=15$x=15 reveal $y$y values of $y=13$y=13, $y=11$y=11, $y=10$y=10, $y=9$y=9 and $y=8$y=8.

On the left hand side, moving away from the asymptote $x=3$x=3 , the values $x=2$x=2, $x=1$x=1, $x=0$x=0, $x=-1$x=−1, $x=-3$x=−3, $x=-9$x=−9 reveal integer $y$y values of $y=-5$y=−5, $y=1$y=1, $y=3$y=3, $y=4$y=4, $y=5$y=5, $y=6$y=6.

The dilation factor $a=12$a=12 has many factors, and we were lucky to pick up $6$6 integer points because of it. In general however the dilation factor will not be as abundant in factors, and so sometimes we are forced to plot a number of non-integer points.

Here is the table of values,  using just $10$10 of the above points, split up to show the points on each side of the asymptote:

Left side $x$x	$-9$−9	$-3$−3	$-1$−1	$0$0	$2$2
$y$y	$6$6	$5$5	$4$4	$3$3	$-5$−5

Right side $x$x	$4$4	$5$5	$7$7	$9$9	$15$15
$y$y	$19$19	$13$13	$10$10	$9$9	$8$8

The $x$x values chosen gives as wide a range as possible so that the imagined arcs might be easier to draw after the points have been plotted. 

Here is the plot. 

 

Practice question

Question 8

 

 

Find the equation of a transformation of $y=\frac{k}{x}$y=kx​

Since every point on a curve satisfies its equation, then we can use this fact to determine the unknown coefficients of a curve's equation.

Worked example

Question 9

Find the equation of the curve that is of the form $y=\frac{k}{x}$y=kx​ and goes through $\left(2,-3\right)$(2,−3).

Do: Suppose we know that $\left(2,-3\right)$(2,−3) is a point on the curve $y=\frac{k}{x}$y=kx​. Then we can argue as follows:

$y$y	$=$=	$\frac{k}{x}$kx​
$-3$−3	$=$=	$\frac{k}{2}$k2​
$k$k	$=$=	$-6$−6

Thus the equation of the curve is $y=-\frac{6}{x}$y=−6x​, and all points on the curve will satisfy it.

Reflect: The curve should be recognized as a transformation of $y=\frac{k}{x}$y=kx​, with a dilation factor of $-6$−6 and possessing mutually perpendicular asymptotes of $x=0$x=0 and $y=0$y=0. The arcs of the graph lie in the upper left and lower right quadrants formed by these asymptotes.

With the equation uniquely defined, we can now easily find other points that satisfy it. For example $\left(1,-6\right)$(1,−6) and $\left(3,-2\right)$(3,−2) are two more points on the curve.

 

Applications involving transformations of  $y=\frac{k}{x}$y=kx​

The following examples may look different, but the fundamental idea of each application is the same. If the product of two variables, say $x$x and $y$y remains constant, then the size of any one of the variables is inversely proportional to the size of the other.

In mathematical terms, if $xy=k$xy=k, then both $y=\frac{k}{x}$y=kx​ and $x=\frac{k}{y}$x=ky​. If $k$k is non-zero, neither $x$x nor $y$y can disappear (become zero). However the variables can become very small or very large, but never so together.

Worked examples

Question 10

A rider has been allowed to have her horses graze on  $400m^2$400m2 of grazing land on a wealthy farmer's extensive property. The father of the rider is willing to build a rectangular enclosure that will contain the horses. The father allows the rider to decide on the dimensions of the enclosure, but insists that the area is no more than that agreed to. What are the riders options, and if the enclosure is to be fenced at $\$35$$35 per meter, how much will it cost the farmer?  

At first, it may seem that the fencing costs will not vary with the dimensions, but it turns out they do.

The basic problem here is to decide the length $l$l and width $w$w given that $l\times w=400$l×w=400. We could set up a table of values for $l$l and see how $w=\frac{400}{l}$w=400l​ varies. Note that $w=\frac{400}{l}$w=400l​ is a transformation of $y=\frac{k}{x}$y=kx​ centered on the origin, but because the length and width are positive numbers, only the arc in the first quadrant is relevant to our problem.

$l$l	$5$5	$10$10	$20$20	$40$40	$80$80
$w=\frac{400}{l}$w=400l​	$80$80	$40$40	$20$20	$10$10	$5$5

We could also determine fencing costs as $C=\left(2l+2w\right)\times35$C=(2l+2w)×35 and see how the costs vary. The second table shows this;

$l$l	$5$5	$10$10	$20$20	$40$40	$80$80
$w=\frac{400}{l}$w=400l​	$80$80	$40$40	$20$20	$10$10	$5$5
$C$C	$5950$5950	$3500$3500	$2800$2800	$3500$3500	$5950$5950

It seems that a square enclosure would be the cheapest option, but there may be other considerations that the rider may need to take into account.

 

Question 11

A drive to Campbelltown from Sam's place is $250$250 km. How would Sam's average speed (in km/h) vary with the time taken to get to Cambelltown if she decided to take a trip tomorrow?

To answer this, we need to know that the speed $s$s in km/h is related to the distance $d$d kilometers and the time $t$t hours by the equation:

$s=\frac{d}{t}$s=dt​

Since the distance is given as the constant $250$250, the average speed, using function notation, can be expressed as:

$s\left(t\right)=\frac{250}{t}$s(t)=250t​

Thus the time varies inversely to the speed of Sam's car, as demonstrated in this graph:

  

The two highlighted points have coordinates $\left(1,250\right)$(1,250) and $\left(5,50\right)$(5,50) determined as $s\left(1\right)=\frac{250}{1}=250$s(1)=2501​=250 km/h and $s\left(5\right)=\frac{250}{5}=50$s(5)=2505​=50 km/h.

If it were possible to travel at a constant speed of $110$110 km/h (the legal speed limit) for the entire journey, we would need to solve for $t$t in the equation $110=\frac{250}{t}$110=250t​:

$110$110	$=$=	$\frac{250}{t}$250t​
$110t$110t	$=$=	$250$250
$t$t	$=$=	$\frac{250}{110}$250110​
$t$t	$=$=	$2.273$2.273

This can interpreted as about $2$2 hours and $16$16 minutes. Of course in practice this would be very hard to achieve.

 

Practice questions

QUESTION 12

QUESTION 13

QUESTION 14