Functions (Graphs and Behaviour)

Hong Kong

Stage 4 - Stage 5

Lesson

The circle centred on the origin with radius $2$2 has the equation $x^2+y^2=4$`x`2+`y`2=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.

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`=`x`2 (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`=`x`2 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 $\frac{x^2}{9}+\frac{y^2}{4}=1$`x`29+`y`24=1, with a translation of $3$3 units to the right and $1$1 unit down. In fact the curve is an ellipse.

A certain curve given by $y=x^2$`y`=`x`2 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`=2`x`−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`=`x`2, 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`=2`x`−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`=2`x`+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=2`x`+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:

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

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

If the graph of $y=-x^2$`y`=−`x`2 is translated horizontally $6$6 units to the right and translated vertically $5$5 units upwards, what is its new equation?

This is a graph of $y=3^x$`y`=3`x`.

How do we shift the graph of $y=3^x$

`y`=3`x`to get the graph of $y=3^x-4$`y`=3`x`−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`=3`x`−4 on the same graph as $y=3^x$`y`=3`x`.

This is a graph of $y=\sqrt{4-x^2}$`y`=√4−`x`2.

Loading Graph...

How do we shift the graph of $y=\sqrt{4-x^2}$

`y`=√4−`x`2 to get the graph of $y=\sqrt{4-x^2}+2$`y`=√4−`x`2+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−`x`2+2 on the same graph as $y=\sqrt{4-x^2}$`y`=√4−`x`2.Loading Graph...