Lesson

Consider the graphs of $y=\sin x$`y`=`s``i``n``x` and $y=-2\sin\left(3x+45^\circ\right)+2$`y`=−2`s``i``n`(3`x`+45°)+2 which are drawn below.

The graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2 |

Starting with the graph of $y=\sin x$`y`=`s``i``n``x`, we can work through a series of transformations so that it coincides with the graph of $y=-2\sin\left(3x+45^\circ\right)+2$`y`=−2`s``i``n`(3`x`+45°)+2.

We can first reflect the graph of $y=\sin x$`y`=`s``i``n``x` about the $x$`x`-axis. This is represented by applying a negative sign to the function (multiplying the function by $-1$−1).

The graph of $y=-\sin x$y=−sinx |

Then we can increase the amplitude of the function to match. This is represented by multiplying the $y$`y`-value of every point on $y=-\sin x$`y`=−`s``i``n``x` by $2$2.

The graph of $y=-2\sin x$y=−2sinx |

Next we can apply the period change that is the result of multiplying the $x$`x`-value inside the function by $3$3. This means that to get a particular $y$`y`-value, we can put in an $x$`x`-value that is $3$3 times smaller than before. Notice that the points on the graph of $y=-2\sin x$`y`=−2`s``i``n``x` move towards the vertical axis by a factor of $3$3 as a result.

The graph of $y=-2\sin3x$y=−2sin3x |

Our next step will be to obtain the graph of $y=-2\sin\left(3x+45^\circ\right)$`y`=−2`s``i``n`(3`x`+45°), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factorise the function into the form $y=-2\sin\left(3\left(x+15^\circ\right)\right)$`y`=−2`s``i``n`(3(`x`+15°)).

In this form, we can see that the $x$`x`-values are increased by $15^\circ$15° inside the function. This means that to get a particular $y$`y`-value, we can put in an $x$`x`-value that is $15^\circ$15° smaller than before. Graphically, this corresponds to shifting the entire function to the **left** by $15^\circ$15°.

The graph of $y=-2\sin\left(3x+45^\circ\right)$y=−2sin(3x+45°) |

Lastly, we translate the graph of $y=-2\sin\left(3x+45^\circ\right)$`y`=−2`s``i``n`(3`x`+45°) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+45^\circ\right)+2$`y`=−2`s``i``n`(3`x`+45°)+2.

The graph of $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2 |

Careful!

When we geometrically apply each transformation to the graph of $y=\sin x$`y`=`s``i``n``x`, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting about the $x$`x`-axis, we would have needed to translate the graph downwards first.

In the example above we were transforming the graph of $y=\sin x$`y`=`s``i``n``x`. The particular function $y=\sin x$`y`=`s``i``n``x` was not important, however. We could have just as easily transformed the graph of $y=\cos x$`y`=`c``o``s``x`, or even a non-trigonometric function, using the same method!

Consider a function $y=f\left(x\right)$`y`=`f`(`x`). Then we can obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d`, where $a,b,c,d$`a`,`b`,`c`,`d` are constants, by applying a series of transformations to the graph of $y=f\left(x\right)$`y`=`f`(`x`). These transformations are summarised below.

Summary

To obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d` from the graph of $y=f\left(x\right)$`y`=`f`(`x`):

- $a$
`a`vertically enlarges the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $b$
`b`horizontally enlarges the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $c$
`c`horizontally translates the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $d$
`d`vertically translates the graph of $y=f\left(x\right)$`y`=`f`(`x`).

In the case that $a$`a` is negative, it has the additional property of reflecting the graph of $y=f\left(x\right)$`y`=`f`(`x`) about the horizontal axis.

If $y=f\left(x\right)$`y`=`f`(`x`) is the equation of a trigonometric function, then a vertical enlargement corresponds to an amplitude change, a horizontal enlargement corresponds to a period change and a horizontal translation corresponds to a phase shift.

The signs of $c$`c` and $d$`d` determine the direction of the horizontal and vertical translations respectively. If $c$`c` is positive the transformation describes a translation to the right, and if $c$`c` is negative the transformation describes a translation to the left. If $d$`d` is positive the transformation describes a translation upwards, and if $d$`d` is negative the transformation describes a translation downwards.

Careful!

If $c$`c` is negative, it may be convenient to represent the equation in the form $y=af\left(b\left(x+c\right)\right)+d$`y`=`a``f`(`b`(`x`+`c`))+`d` instead, where we've redefined $c$`c` using its absolute value. In this case, the value of $c$`c` represents translation to the left.

Similarly, if $d$`d` is negative, it may be convenient to represent the equation in the form $y=af\left(b\left(x-c\right)\right)-d$`y`=`a``f`(`b`(`x`−`c`))−`d`, where we've redefined $d$`d` using its absolute value. In this case, the value of $d$`d` represents translation downwards.

Lastly, the magnitude of $a$`a` and $b$`b` determine whether the vertical and horizontal dilations each describe a compression or an expansion.

For a value of $a$`a` where $\left|a\right|>1$|`a`|>1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) vertically expands or stretches. For a trigonometric function, we say that the amplitude increases. If $\left|a\right|<1$|`a`|<1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) vertically compresses. For a trigonometric function, we say that the amplitude decreases.

For a value of $b$`b` where $\left|b\right|>1$|`b`|>1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) horizontally compresses. If $\left|b\right|<1$|`b`|<1, then the graph horizontally expands or stretches. In the case that the graph describes a trigonometric function, a horizontal compression means the period decreases and a horizontal expansion means the period increases.

Consider the graphs of $y=\sin x$`y`=`s``i``n``x` and $y=5\sin\left(x+\left(\left(-60\right)\right)\right)$`y`=5`s``i``n`(`x`+((−60))).

Loading Graph...

What transformations have occurred?

Select all that apply.

Vertical translation

AHorizontal translation

BVertical enlargement

CHorizontal enlargement

DVertical translation

AHorizontal translation

BVertical enlargement

CHorizontal enlargement

DComplete the following statement.

The graph of $y=\sin x$

`y`=`s``i``n``x`has increased its amplitude by a factor of $\editable{}$ units and has undergone a phase shift of $\editable{}$ to the right.

The graph of $y=\cos x$`y`=`c``o``s``x` has been transformed into the graph of $y=\cos\left(2x+\left(\left(-60\right)\right)\right)$`y`=`c``o``s`(2`x`+((−60))).

What transformations have occurred?

Select all that apply.

Vertical translation

AHorizontal translation

BHorizontal enlargement

CVertical enlargement

DVertical translation

AHorizontal translation

BHorizontal enlargement

CVertical enlargement

DComplete the following statement.

The graph of $y=\cos x$

`y`=`c``o``s``x`has decreased its period by a factor of $\editable{}$ and then has undergone a phase shift of $\editable{}$ to the right.Draw the graph of $y=\cos\left(2x+\left(\left(-60\right)\right)\right)$

`y`=`c``o``s`(2`x`+((−60))).Loading Graph...

The graph of $y=\sin x$`y`=`s``i``n``x` undergoes the series of transformations below.

What is the equation of the transformed graph in the form $y=-\sin\left(x+c\right)+d$`y`=−`s``i``n`(`x`+`c`)+`d` where $c$`c` is the lowest positive value in degree?

- The graph is reflected about the $x$
`x`-axis. - The graph is horizontally translated to the left by $60^\circ$60°.
- The graph is vertically translated downwards by $3$3 units.