Trigonometric Graphs

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Find the equation of a tangent curve

Lesson

The graph of the function defined by the expression $\tan x$`t``a``n``x` is transformed in various ways when constants $a$`a`, $b$`b`, $c$`c` and $d$`d` are chosen to make a new function definition, $y(x)=a\tan(bx+c)+d$`y`(`x`)=`a``t``a``n`(`b``x`+`c`)+`d`.

The effects of these constants are the same as in the corresponding transformations of the sine and cosine functions.

The constant $a$`a` multiplies every function value. Therefore, if $|a|>1$|`a`|>1, the graph appears stretched vertically, and if $0<|a|<1$0<|`a`|<1, the graph is compressed vertically. The *absolute value* of $a$`a` has been used because when $a$`a` is negative, the graph not only shows the dilation effect but it also appears reflected across the horizontal axis.

The constant $b$`b` that multiplies the variable $x$`x`, affects the horizontal scale. It does this in a way that may be unexpected: For $|b|>1$|`b`|>1, the graph appears compressed towards the origin by the factor $\frac{1}{b}$1`b`, and for $|b|<1$|`b`|<1, the stretching is away from the origin by the factor $\frac{1}{b}$1`b`. As before, we have used the absolute value symbol because if $b$`b` is negative, there is a reflection across the vertical axis as well as the dilation.

The consequence for the $\tan$`t``a``n` function (and the other trigonometric functions) is that $b$`b` affects the period of a periodic function. For example, $\tan x$`t``a``n``x` has period $\pi$π but $\tan2x$`t``a``n`2`x` has period $\frac{1}{2}\pi$12π. Similarly, in degrees, $\tan\theta$`t``a``n``θ` has period $180^\circ$180° but $\tan\frac{\theta}{2}$`t``a``n``θ`2 has period $180^\circ\times\frac{1}{\frac{1}{2}}=360^\circ$180°×112=360°.

The constant $c$`c` added to $bx$`b``x` brings about a horizontal translation of the graph. The amount of the translation or sideways shift is equal to the amount that has been added to $x$`x` directly. If the argument of the function is $bx+c$`b``x`+`c`, we must think of this as $b(x+\frac{c}{b})$`b`(`x`+`c``b`) to see that the amount added directly to $x$`x` is $\frac{c}{b}$`c``b`, and this is the sideways shift.

If $c$`c` is positive, the shift is to the left and if $c$`c` is negative, the shift is to the right.

The constant $d$`d` is added to every function value. Therefore, it has the effect of translating the graph vertically. The amount of the translation is $d$`d` itself.

If we are given a graph that has the shape of a $\tan$`t``a``n` function graph, and we wish to produce its equation with the correct constants $a$`a`, $b$`b`, $c$`c` and $d$`d`, we must use the facts about transformations that have been explained above.

Displayed below is the graph of $\tan x$`t``a``n``x` between $-\frac{\pi}{2}$−π2 and $2\pi$2π. Here, $a=1$`a`=1, $b=1$`b`=1, $c=0$`c`=0 and $d=0$`d`=0. We will use this for comparison with other graphs involving $\tan$`t``a``n`.

Shown below is the graph (coloured green) of a function involving $\tan$`t``a``n`. It is superimposed on the graph of $\tan x$`t``a``n``x` between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2. We wish to find the correct constants for its equation.

We see that three periods of the green curve occupy the same horizontal space as one period of $\tan x$`t``a``n``x`. The period has been multiplied by the factor $\frac{1}{3}$13. Equivalently, we say that the horizontal scale has been reduced by the factor $\frac{1}{3}$13.

No other transformation is apparent. So, we conclude that the equation of the green curve is $y(x)=\tan3x$`y`(`x`)=`t``a``n`3`x`.

Shown below is the graph (coloured green) of another function involving $\tan$`t``a``n`. As before, it is superimposed on the graph of $\tan x$`t``a``n``x` between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2. We wish to find the correct constants for its equation.

The unknown graph appears to have the same centre and the same period as the graph of $\tan x$`t``a``n``x`. We note that while $\tan\frac{\pi}{4}=1$`t``a``n`π4=1 as shown, the green curve has the value $3$3 at $\frac{\pi}{4}$π4. In fact, at every point on the $x$`x`-axis, the green curve has $3$3 times the height of the $\tan$`t``a``n` curve. There has been a vertical dilation by the factor $3$3. Therefore, the green curve is the graph of $y(x)=3\tan x$`y`(`x`)=3`t``a``n``x`.

This example has several transformations of the $\tan$`t``a``n` function combined. As before, we are looking for the constants $a$`a`, $b$`b`, $c$`c` and $d$`d`.

The first step should be to find the centre line of the unknown graph. Judging exactly where it is is not easy but we locate it so that there will be rotational symmetry about a point where the graph intersects the proposed centre line. In this example, the line is at $y=1$`y`=1 and we conclude that the graph has been translated vertically by $1$1 unit. That is, $d=1$`d`=1.

By drawing a vertical line through the nearest intersection point with the centre line to the $y$`y`-axis we can see that the graph has been shifted to the right by a little over $0.5$0.5 and with very careful measurement we might conclude that the shift is, in fact, $\frac{\pi}{6}$π6.

The period of the green curve is the same as the period of the $\tan$`t``a``n` curve. So, $b=1$`b`=1.

Now, $\frac{c}{b}=\frac{\pi}{6}$`c``b`=π6. So, we have $c=\frac{\pi}{6}$`c`=π6.

Finally, we need to find the constant $a$`a`. This is not a perfectly easy task. There are several strategies you could try and the following are two suggestions.

We know that $\tan\frac{\pi}{4}=1$`t``a``n`π4=1 and that the green curve has been moved to the right by $\frac{\pi}{6}$π6 and up by $1$1. We need to shift both curves so that their centres are aligned in order to compare their vertical scales.

If the $\tan$`t``a``n` curve were to be moved to the right by $\frac{\pi}{6}$π6, it would attain the value $1$1 at $\frac{\pi}{6}+\frac{\pi}{4}=\frac{5\pi}{12}\approx1.31$π6+π4=5π12≈1.31.

Reading from the graph, we see that the green curve attains the value $3$3 at $1.31$1.31, and so, if the green curve were to be moved down by $1$1 unit to align its centre line with that of the $\tan$`t``a``n` curve, it would attain the value $2$2 at $1.31$1.31. This is $2\times1$2×1. So, the vertical dilation must be by a factor of $2$2 and therefore, $a=2$`a`=2.

We can see from the graphs that the green curve and the $\tan x$`t``a``n``x` curve intersect at several points. That is, their values are equal at those points. Using the values of $b$`b`, $c$`c` and $d$`d` that we have determined, we can write $a\tan(x-\frac{\pi}{6})+1=\tan x$`a``t``a``n`(`x`−π6)+1=`t``a``n``x` for $x$`x` at any one of the intersection points.

One such point is approximately $x=4.35$`x`=4.35 (reading from the graph). If this value is substituted for $x$`x` in the equation just derived, we can, with the help of a calculator, solve for $a$`a`.

$a=\frac{\tan4.35-1}{\tan(4.35-\frac{\pi}{6})}\approx2.005$`a`=`t``a``n`4.35−1`t``a``n`(4.35−π6)≈2.005

We now have the equation for the green curve:

$y(x)=2\tan(x-\frac{\pi}{6})+1$`y`(`x`)=2`t``a``n`(`x`−π6)+1

Consider the graph of the given $\tan$`t``a``n` function.

Loading Graph...

The given graph is the result of what transformation?

Translating the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.ATranslating the graph of $y=\tan x$

`y`=`t``a``n``x`vertically.BReflecting the graph of $y=\tan x$

`y`=`t``a``n``x`about the $x$`x`-axis.CTranslating the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.ATranslating the graph of $y=\tan x$

`y`=`t``a``n``x`vertically.BReflecting the graph of $y=\tan x$

`y`=`t``a``n``x`about the $x$`x`-axis.CState the equation of the graphed function.

Consider the graph of the given $\tan$`t``a``n` function.

Loading Graph...

The given graph is the result of what transformation?

Translating the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.ACompressing the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.BReflecting the graph of $y=\tan x$

`y`=`t``a``n``x`about the $x$`x`-axis.CDilating the graph of $y=\tan x$

`y`=`t``a``n``x`vertically.DTranslating the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.ACompressing the graph of $y=\tan x$

`y`=`t``a``n``x`horizontally.BReflecting the graph of $y=\tan x$

`y`=`t``a``n``x`about the $x$`x`-axis.CDilating the graph of $y=\tan x$

`y`=`t``a``n``x`vertically.DState the equation of the graphed function.

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions