Amplitude, period and phase shift of can be recovered from the graph by noticing the coordinates of the peaks and troughs of the wave.
Let $y_{peak}$ and $y_{trough}$ denote the $y$-coordinates of the peaks and troughs of the wave. Then for the amplitude we have
$$ A=\frac{1}{2}\left(y_{peak}-y_{trough}\right) $$
From your graph there is no indication of the vertical scale, but let us suppose that the horizontal dashed lines are one unit apart. Then we would have $y_{peak}=2$ and $y_{trough}=-6$. This would give $A=\frac{1}{2}(2-(-6))=4$.
We can also use $y_{peak}$ and $y_{trough}$ to find the vertical shift $D$.
$$ D=\frac{1}{2}\left(y_{peak}+y_{trough}\right) $$
So, for your example, $D=\frac{1}{2}(2+(-6))=-2$.
This leaves the values of $B$ and $C$. But first, we must find the period $P$, which is straightforward.
The period P is the horizontal distance between two successive peaks of the graph. For your graph this would be a distance $P=3\pi$.
The value of $B$ is then found from
$$ B=\frac{2\pi}{P} $$
For your graph, then, we have $B=\frac{2\pi}{3\pi}=\frac{2}{3}$.
Finally we have the phase shift $\phi$.
The phase shift for the sine and cosine are computed differently. The easiest to determine from the graph is the phase shift of the cosine.
Let $x_{peak}$ denote the $x$-coordinate of the peak closest to the vertical axis. For your graph, we have $x_{peak}=0$.
$$\phi=x_{peak}\text{ for the cosine graph}$$
$$\phi=\left(x_{peak}-\frac{P}{4}\right)\text{ for the sine graph}$$
For your graph this gives phase shift $\phi=0$ for the cosine graph and phase shift $\phi=0-\frac{3\pi}{4}=-\frac{3\pi}{4}$ for the sine graph.
But for your equations, we need the value of $C$. The value of $C$ is found from the values of $\phi$ and $B$.
The equation for $C$ in terms of the phase shift $\phi$ is
$$ C=B\phi $$
So if we use the cosine function to model your graph we have $C=0$ and for the sine graph we have $C=\frac{2}{3}\cdot\left(-\frac{3\pi}{4}\right)=-\frac{\pi}{2}$.
So we have for both sine and cosine
- $A=4$
- $D=-2$
- $B=\frac{2}{3}$
For cosine, $C=0$ and for sine, $C=-\frac{\pi}{2}$.
Thus your graph can be represented by either of the two equations
$$ y=4\cos\left(\frac{2}{3}x \right)-2 $$
$$ y=4\sin\left(\frac{2}{3}x+\frac{\pi}{2}\right)-2 $$
We are looking for the maximum of $$g(x)=\frac {x^4-x}{x^2+1} - (x^2-1) = \frac {x^4-x}{x^2+1} -\frac{x^4-1}{x^2+1}=\frac {1-x}{1+x^2}.$$
To find critical points, set $$g'(x)=\frac {-(1+x^2)-(1-x)2x}{(1+x^2)^2}=\frac{-1-2x+x^2}{(1+x^2)^2}=0.$$
Setting the numerator equal to zero and solving the quadratic equation yields $x=\frac{2\pm \sqrt{8}}2=1\pm \sqrt2.$
Note that $g(0)=1$ and $g(1)=0$ and $1-\sqrt2<0<1<1+\sqrt2,$
so $g(1-\sqrt2)$ is a maximum and $g(1+\sqrt2)$ is a minimum.
The maximum value is thus $g(1-\sqrt2)=\frac{\sqrt2}{4-2\sqrt2}=\frac {\sqrt2+1}2=1.207...$
Best Answer
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=\frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+\frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
