Translations of relations
The circle centred on the origin with radius $2$2 has the equation $x^2+y^2=4$x2+y2=4.
Suppose we replace the $x$x with $\left(x-15\right)$(x−15) and the $y$y with $\left(y-9\right)$(y−9) so that our new equation becomes $\left(x-15\right)^2+\left(y-9\right)^2=4$(x−15)2+(y−9)2=4. The two relations are depicted here.
The change from $x$x to $\left(x-15\right)$(x−15) and from $y$y to $\left(y-9\right)$(y−9) caused the circle to shift to the right by $15$15 units and upward by $9$9 units. The centre moved from $\left(0,0\right)$(0,0) to $\left(15,9\right)$(15,9) and the radius (and thus the overall shape of the circle) remained unchanged.
We can think of the two movements (right and up) as happening independently. The horizontal translation of $15$15 units and the vertical translation of $9$9 units.
Understanding a translated form
When we think about a function like $y=\left(x+3\right)^2-5$y=(x+3)2−5, we may consider it as the translation of the basic function $y=x^2$y=x2 (a parabola sitting upright with its minimum turning point at the origin). This means that $y=\left(x+3\right)^2-5$y=(x+3)2−5, which could be re-written $\left(y+5\right)=\left(x+3\right)^2$(y+5)=(x+3)2, is simply $y=x^2$y=x2 translated $3$3 units to the left, and $5$5 units down.
Something as straight-forward as the linear function $y=x-2$y=x−2 can be thought of as the line $y=x$y=x shifted to the right $2$2 units. It could equally be thought of as $y=x$y=x shifted downward by $2$2 units. Can you see why?
Consider more complicated relations like $\frac{\left(x-3\right)^2}{9}+\frac{\left(y+1\right)^2}{4}=1$(x−3)29+(y+1)24=1. While we perhaps maybe uncertain about what its basic shape may look like, we understand that its essential form is given by $\frac{x^2}{9}+\frac{y^2}{4}=1$x29+y24=1, with a translation of $3$3 units to the right and $1$1 unit down. In fact the curve is an ellipse.
An interesting example
A certain curve given by $y=x^2$y=x2 is translated $1$1 unit to the left and $3$3 units upward. In addition to this, the line given by $y=2x-1$y=2x−1 is translated $3$3 units to the left. Where do these translated graphs intersect each other?
We first need to understand that the parabola $y=x^2$y=x2, after translation, will have the form $y-3=\left(x+1\right)^2$y−3=(x+1)2. We can write this as $y=\left(x+1\right)^2+3$y=(x+1)2+3. We also see that the line $y=2x-1$y=2x−1, after translation, becomes $y=2\left(x+3\right)-1$y=2(x+3)−1. After simplification this line has the equation $y=2x+5$y=2x+5.
At the intersections of these translated graphs, the $y$y - values are equal, so we can put $\left(x+1\right)^2+3=2x+5$(x+1)2+3=2x+5 and solve for $x$x.
| $\left(x+1\right)^2+3$(x+1)2+3 | $=$= | $2x+5$2x+5 |
| $x^2+2x+4$x2+2x+4 | $=$= | $2x+5$2x+5 |
| $x^2-1$x2−1 | $=$= | $0$0 |
| $\left(x-1\right)\left(x+1\right)$(x−1)(x+1) | $=$= | $0$0 |
| $x$x | $=$= | $\pm1$±1 |
At $x=-1$x=−1, $y=3$y=3 and at $x=1$x=1, $y=7$y=7.
The translated graphs and the points of intersection are depicted here:
Worked Examples
Question 1
How do we shift the graph of $y=f\left(x\right)$y=f(x) to get the graph of $y=f\left(x\right)+4$y=f(x)+4?
Move the graph up by $4$4 units.
AMove the graph down by $4$4 units.
B
Question 2
How do we shift the graph of $y=g\left(x\right)$y=g(x) to get the graph of $y=g\left(x+6\right)$y=g(x+6)?
Move the graph to the left by $6$6 units.
AMove the graph to the right by $6$6 units.
B
Question 3
If the graph of $y=-x^2$y=−x2 is translated horizontally $6$6 units to the right and translated vertically $5$5 units upwards, what is its new equation?
Question 4
This is a graph of $y=3^x$y=3x.
How do we shift the graph of $y=3^x$y=3x to get the graph of $y=3^x-4$y=3x−4?
Move the graph $4$4 units to the right.
AMove the graph downwards by $4$4 units.
BMove the graph $4$4 units to the left.
CMove the graph upwards by $4$4 units.
DHence, plot $y=3^x-4$y=3x−4 on the same graph as $y=3^x$y=3x.
Loading Graph...A number plane with the exponential function y=a^x plotted.
Question 5
This is a graph of $y=\sqrt{4-x^2}$y=√4−x2.
How do we shift the graph of $y=\sqrt{4-x^2}$y=√4−x2 to get the graph of $y=\sqrt{4-x^2}+2$y=√4−x2+2?
Move the graph to the right by $2$2 units.
AMove the graph to the left by $2$2 units.
BMove the graph downwards by $2$2 units.
CMove the graph upwards by $2$2 units.
DHence plot $y=\sqrt{4-x^2}+2$y=√4−x2+2 on the same graph as $y=\sqrt{4-x^2}$y=√4−x2.
Loading Graph...