Principles of translation
The hyperbola given by $y=\frac{a}{x-h}+k$y=ax−h+k can be thought of as the basic rectangular hyperbola $y=\frac{a}{x}$y=ax translated horizontally (parallel to the $x$x axis) a distance of $h$h units and translated vertically (parallel to the $y$y axis) a distance of $k$k units. We note that under this type of translation:
The centre will move to the point $\left(h,k\right)$(h,k).
The orientation of the hyperbola will remain unaltered.
The asymptotes will become the straight lines $x=h$x=h and $y=k$y=k.
For example, to sketch the hyperbola $y=\frac{12}{x-3}+7$y=12x−3+7, first place the centre at $\left(3,7\right)$(3,7). Then draw in the two orthogonal asymptotes (orthogonal means at right angles) given by $x=3$x=3 and $y=7$y=7. Finally, draw the hyperbola as if it were the basic hyperbola $y=\frac{12}{x}$y=12x but now centred on the point $\left(3,7\right)$(3,7).
Note that the domain includes all values of $x$x not equal to $3$3 and the range includes all values of $y$y not equal to $7$7. Here is a graph showing how the basic function $y=\frac{12}{x}$y=12x is translated to horizontally and vertically to become the transformed function $y=\frac{12}{x-3}+7$y=12x−3+7.
We can also note that as $x\rightarrow3,y\rightarrow\infty$x→3,y→∞and as $x\rightarrow\infty,y\rightarrow7$x→∞,y→7.
Finding points
Continuing with our example, to find the point where $x=9$x=9, we simply substitute $x=9$x=9 into $y=\frac{12}{x-3}+7$y=12x−3+7 so that $y=\frac{12}{9-3}+7$y=129−3+7 or when simplified $y=9$y=9. Thus the point $\left(9,9\right)$(9,9) lies on the hyperbola.
To find the point where $y=13$y=13, set $13=\frac{12}{x-3}+7$13=12x−3+7 and solve for $x$x, so that:
| $13$13 | $=$= | $\frac{12}{x-3}+7$12x−3+7 |
| $6$6 | $=$= | $\frac{12}{x-3}$12x−3 |
| $6\left(x-3\right)$6(x−3) | $=$= | $12$12 |
| $6x-18$6x−18 | $=$= | $12$12 |
| $6x$6x | $=$= | $30$30 |
| $x$x | $=$= | $5$5 |
Thus another point on the hyperbola is $\left(5,13\right)$(5,13).
Applet
Use the applet to understand how the translation of the basic function works. Try positive and negative values of $a$a.
Worked Examples
QUESTION 1
This is a graph of the hyperbola $y=\frac{1}{x}$y=1x.
A Cartesian plane has an $x$x-axis and $y$y-axis ranging from $-10$−10 to $10$10. Each axis has major tick marks at $2$2-unit intervals and minor tick marks at $1$1-unit intervals. A graph of a hyperbola $y=\frac{1}{x}$y=1x with two branches is plotted. One branch lies in the first quadrant, and the other branch lies in the third quadrant.
What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted upwards by $4$4 units?
What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted to the right by $7$7 units?
QUESTION 2
This is a graph of $y=\frac{1}{x}$y=1x.
How do we shift the graph of $y=\frac{1}{x}$y=1x to get the graph of $y=\frac{1}{x}+3$y=1x+3?
Move the graph $3$3 units to the left.
AMove the graph upwards by $3$3 unit(s).
BMove the graph downwards by $3$3 unit(s).
CMove the graph $3$3 units to the right.
DHence sketch $y=\frac{1}{x}+3$y=1x+3 on the same graph as $y=\frac{1}{x}$y=1x.
Loading Graph...A cartesian plane is shown with both the x-axis and y-axis ranging from -10 to 10. A hyperbola is drawn with a function y = 1/x.
QUESTION 3
Answer the following.
Consider $y=\frac{-1}{x}$y=−1x. What value cannot be substituted for $x$x?
In which two quadrants does the graph of $y=\frac{-1}{x}$y=−1x lie?
1
A2
B4
C3
DConsider $y=\frac{-1}{x-4}$y=−1x−4. What value cannot be substituted for $x$x?
In which three quadrants does the graph of $y=\frac{-1}{x-4}$y=−1x−4 lie?
4
A1
B3
C2
DHow can the graph of $y=\frac{-1}{x}$y=−1x be altered to create the graph of $y=\frac{-1}{x-4}$y=−1x−4?
translated $4$4 units right
Areflected about $x$x-axis
Bsteepened
Ctranslated $4$4 units down
D