How do I determine the value of $d$ from the tangent graph of the form $y = a\times \tan (bx + c) + d$?

Question

I understand that $d$ is the vertical shift, and I have already worked out the values of $a$, $b$ and $c$ to be the following (please tell me if I am wrong):

$\pmb{a}\quad$ Given point is $(-1, -7)$ so I substituted it in as the values of $x$ and $y$ ; $-7= a\times \tan\left(\dfrac{\pi}{4}\left(-1\right)+5\right)$ which I solved for $a$ and that gave me$\space$ -3.8040.

$\pmb{b}\quad$ Period so $\dfrac{\pi}{4}$

$\pmb{c}\quad$ $y$-intercept so $5$

But how do I find $d$ for this type of graph with only vertical asymptotes?

Answer 1

You are correct in that $b = \frac{\pi}{4}$, but your reasoning isn't so clear. The function's period is $4$, and since $f(x) = \tan (\alpha x)$ has period $\pi / \alpha$, $b$ is $\frac{\pi}{4}$ by simple arithmetic.

So far we have $f(x) = a\tan(\frac{\pi}{4}x + c) + d$.

The function $g(x) = \tan(\frac{\pi}{4}x)$ has asymptotes at $x = 4n + 2$ e.g $x = 2$ where $n$ is an integer, while $f(x)$ has asymptotes at $x = 4n +1$. This means in the simplest case that $f(x)$ is $\tan(\frac{\pi}{4}x)$ translated $1$ unit to the left.

Hence $$ f(x) = a\tan\left(\frac{\pi}{4}(x+1)\right) + d \tag{1}\label{eq1}$$

Substitute the value $f(-1) = -7$ into \eqref{eq1} to obtain that $d = -7$, and substitute $f(0) = -5$ to obtain $a = 2$.

Finally we arrive at $$f(x) = 2 \tan \left(\frac{\pi}{4}x + \frac{\pi}{4}\right) -7$$

I hope this answer was clear.

This is my first time giving an answer on this website, if you are a more experienced user and have any constructive criticisms feel free to share them with me.

Answer 2

The graph of a transformed $\tan$ function is

$ f(x) = A \tan( B (x - C) ) + D $

The first thing is that $ B = \dfrac{\pi}{T} $ where $T $ is the period.

Here the period is $ 4 $ , so $B = \dfrac{\pi}{4} $

$C $ is the horizontal shift to the right. Here the graph is shifted to the left by $1$, therefore, C = -1 $

Now we have two points on the graph: $ (-1, -7)$ and $(0, -5) $. Hence

$ -7 = 0 + D $ from which $D = -7 $

and

$ -5 = A \tan( \dfrac{\pi}{4} ( 0 + 1 ) ) - 7 $

From which $ A = 2 $

Therefore, the function is

$ f(x) = 2 \tan( \dfrac{\pi}{4} x + \dfrac{\pi}{4} ) - 7 $