Function for drawing $\sin x$ with a maximum value of $360$ and minimum of $70$? I want a function to draw $\sin x$ between an upper limit of $360$ and a lower limit of $70$.
Attempt: I know how to handle the lower limit but I'm having trouble with the upper limit.
$$y = a(\sin(x)+1) + 70$$
 A: Since $\sin x$ goes from $-1$ to $+1$ you only need to make
$$f(x)=a\sin x+b$$ and solve the system $$\begin{cases}a+b=360\\-a+b=70\end{cases}$$ hence $$\color{red}{(a,b)=(145,215)}$$
A: I assume you want to find some shifted and scaled/stretched/compression version of $\sin x$ such that the upper limit is $360$ and the lower limit is $70$.
Then halfway between $y=360$ and $y=70$, we must have the midline of $y=215$. From $215$ we need to go up by an amplitude of $360-215=145$  to get to $360$ and down by $215-70=145$ to get to $70$. Hence, if we assume the point $(0,215)$, the equation your looking for is:
$$y=145\sin(x)+215$$
This follows from $-1 \leq \sin x \leq 1$, because then $145(1)+215$ will be our max and $145(-1)+215$ will be our minimum. You can add a horizontal stretch/compression if you wish.
If we have the more general point $(0,y(0))$ then using the same idea about the maximum value $\sin (u)$ to assume what you want is of the form:
$$y=145\sin(ax+b)+215$$
This assumption was made from begging (we want $\sin x$ stretched/compressed/shifted in such a way...). Anyways if you plug in $x=0$ you get:
$$c=145\sin(b)+215$$
You can solve for $b$. $a$ can be anything but if you don't want a horizontal stretch from our parent function $\sin x$, take $a=1$ because $a$ "scales" or stretches/ compresses your function in the horizontal direction.
A: Figured it out seconds after posting by luck. :)
$$y= \frac{290}{2}(\sin(x)+1)+70,$$
since $\sin (x)$ occupies a total height of $2$.
