Trigonometric Graphs

New Zealand

Level 7 - NCEA Level 2

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 dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $b$
`b`horizontally dilates 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 dilation corresponds to an amplitude change, a horizontal dilation 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))).

What transformations have occurred?

Select all that apply.

Vertical translation

AHorizontal translation

BVertical dilation

CHorizontal dilation

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 dilation

CVertical dilation

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.