I’m a senior in high school and I’m having trouble understanding how to get the range of a given equation. Example:
$$ A)\quad y = 5\sin(6x + 120°)-6$$
$$ B)\quad y = 0.5\cos(2x -6)+10$$
3 Answers
Use the amplitude and vertical translation to determine the range.
$$y = \color{blue}{a}\sin b(x-h)+\color{purple}{k}\quad y = \color{blue}{a}\cos b(x-h)+\color{purple}{k}$$
In such sinusoidal functions, $\color{blue}{\vert a\vert}$ determines the amplitude and $\color{purple}{k}$ represents the vertical translation. For instance, look at your first example.
$$y = 5\sin(6x+120°)-6$$
It is apparent what $\color{blue}{\vert a\vert}$ and $\color{purple}{k}$ are.
$$\color{blue}{\vert a\vert = 5}$$
$$\color{purple}{k = -6}$$
Amplitude tells us how many units up and down a sine wave oscillates from its point of equilibrium, meaning the graph oscillates between $a$ and $-a$. For $\color{purple}{k = 0}$, the point of equilibrium would be $0$, and the graph would oscillate between $5$ and $-5$.
$$\color{blue}{Range: [-5, 5]}$$
Here, however, $\color{purple}{k = -6}$, so everything is shifted $6$ units down. The point of equilibrium becomes $-5$, and the graph now oscillates between $-1$ and $-11$.
$$\color{blue}{Range: [-5\color{red}{-6}, 5\color{red}{-6}]} \implies \boxed{\color{blue}{Range: [-11, -1]}}$$
Hence, the range of the function is $[-11, -1]$.
You can apply the same logic to any other sine or cosine function.
Also, keep in mind that $h$ affects only horizontal translation, so it has no impact on the range, meaning that different values of $x$ can now produce the same output/range.
As an example, $\sin (x-1°)$ has the exact same range as $\sin x$, yet it’s shifted $1$ unit right.
Let's take apart the first one step-by-step.
- Each value of $x$ on the right produces a unique value for $y$, so we can think of your relation as a function $y=f(x)$ with $f(x) = 5 \sin(6x + 120^\circ)-6$. We now need to determine the range of this function $f$.
- Note that $6x+120^\circ$ generates the entire range of $[0^\circ,360^\circ]$ for different values of $x$. Hence, $S(x) =\sin(6x + 120^\circ)$ can be anything between $-1$ and $1$, which we denote by $[-1,1]$.
- Then, the range of $5S(x)$ is just the range of $S$ scaled (stretched) 5 times, i.e. $[-1\cdot 5,1\cdot 5] = [-5,5]$.
- Finally, the range of $f(x) = 5S(x) - 6$ just shifts the range 6 units down, i.e. you get $$[-1 \cdot 5 -6, 1\cdot 5 - 6] = [-5-6,5-6] = [-11,-1].$$
Can you now do the second one?
To put it in a simple way:
$$ f_1:x\mapsto 6x+120, \mbox{ maps } \color{red}{\Bbb{R}} \mbox{ to } \color{blue}{\Bbb{R}}$$ $$ f_2:x\mapsto \sin(x), \mbox{ maps } \color{blue}{\Bbb{R}} \mbox{ to } \color{purple}{[-1,1]}$$ $$ f_3: x\mapsto 5x-6, \mbox{ maps } \color{purple}{[-1,1]} \mbox{ to } \color{green}{[-11,-1]}$$ Since $$ f(x) = 5\sin(6x + 120)-6=f_3(f_2(f_1(x))))$$ then it $\mbox{ maps } \color{red}{\Bbb{R}} \mbox{ to } \color{green}{[-11,-1]}$ Hence the range of $f$ is $[-11,-1]$
Use the same logic to find the range of the second function.
