How do I sketch $y = -2\sin\big(2(x+\frac{π}{6})\big)+1$ between $[-π,π]$ without any fuss?

Question

I've looked at a bunch of videos on youtube, asked multiple teachers and tutors yet everyone just kind of vaguely told me stuff to do and I didn't understand at all. The parts that mess me up is finding the $x$-intercepts between the domain (especially in the negative section), fitting the graph onto an axis with a scale, and determining the end shapes. I'd like an in-depth explanation on how to sketch this step by step, that I can follow and be successful with other graphs as well.

Answer 1

$$y=-2\sin2(x+\pi/6)+1=-2\sin(2x+\pi/3)+1$$ This is a linear transformation of a sine curve, so the graph will also be sinusoidal, with period $\pi$ because of the $2x$.

The extrema of this curve are obtained by substituting $\pm1$ for the sine, since those are the raw sine function's extrema. We get $3$ as the maximum when the sine is $-1$, and $-1$ as the minimum when the sine is $+1$.

The last thing we need is a place where $y$ attains a maximum, which means $\sin(2x+\pi/3)=-1$ or $2x+\pi/3=-\pi/2$ or $x=-5\pi/12$.

Now to the actual plotting. We place a point at $(-5\pi/12,3)$. Advance a quarter period ($\pi/4$) and place a point midway between the lines $y=-1$ and $y=3$ – that is, at $(-\pi/6,1)$. Advance another quarter period and place a point at the minimum of $y$, i.e. $(\pi/12,-1)$. Now draw a sine curve between these three points; the rest of the graph can be obtained by repeatedly flipping this segment left and right.

Answer 2

1) determining the end shapes

Let's start with $\sin x$. We can get the original function through a series of transformations:

$$\sin(x + \frac{\pi}{6}) \text{ : shift $\sin x$ by $\frac{\pi}{6}$ units to the left}$$ $$\sin(2(x + \frac{\pi}{6})) \text{ : squish the x-axis by a factor of $2$} $$ $$-2\sin(2(x + \frac{\pi}{6})) \text{ : stretch the y-axis by a factor of $-2$} $$ $$-2\sin(2(x + \frac{\pi}{6}))+1 \text{ : shift the whole function upwards by $1$ unit} $$

Therefore, you can just follow these steps to transform the minimum and maximum points of $\sin x$ into the original graph. Since the $x$-axis is squished by a factor of $2$, you will need the points in the range $[-2\pi, 2\pi]$.

Parcly Taxel's answer gives a nice method on how to find the roots, so you can transform those as well.

The key point here is just to find the key points on the graph of $\sin x$ (or $\cos x$, $\tan x$, whichever graph you're using), and then transform them. The end shapes will come out from the minimum and maximum points.

2) fitting the graph onto an axis with a scale

The maximum and minimum points of $\sin x$ are $1$ and $-1$, so the maximum and minimum points of this function are $-2(-1) + 1 = 3$ and $-2(1)+1 = -1$. So your graph needs to span $4$ units on the $y$-axis, and $2 \pi$ units on the $x$-axis.

Let's say your graph paper is $20 \ \text{cm}$ wide and $25 \ \text{cm}$ long. Then for the $y$-axis, your scale is $1 \ \text{cm} : 0.2 \ \text{units}$.

As for the $x$-axis, the largest multiple of $2\pi$ we can get without going over $25$ is $3$. Therefore, the scale would be ${6 \pi} \ \text{cm}: \pi \ \text{units}$ which is just ${1} \text{cm}: \pi/6 \ \text{units}$.