Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$

I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
 A: 
How did you get $b=2k\pi$?

Put $x=\dfrac{\pi}{2}$ in the equation.
$$\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1)$$
Since $\arcsin(1)=\dfrac{\pi}{2}$
You'll get $b+\dfrac{\pi}{4}=k\pi+
(-1)^k\dfrac{\pi}{2}$
Since we have added $b$ in the argument of sine, that is $2b$ in the $x$,  the graph has been shifted backwards by $2b$ units.
Now put $k=1$ , you'll get $b=\dfrac{\pi}{4}$You can try $k=0$ and $k=2$ also, but that all will result in a value of $b$ which isn't in the range $(0,\pi/2)$
(Since any other value of $k$ will lead $b$ to get out of the range $(0,\pi/2)$)
Alternatively since the first negative $x$-intercept of graph is $2b$, and it is clearly visible that it is $\pi/2$, we can say that $$2b=\dfrac{\pi}{2} \implies b=\dfrac{\pi}{4}$$
A: Note that $y=2 \sin \left( \frac{x}{2} \right)$ passes through the origin, and increases to its first peak at $x=\pi$.  
The function $2 \sin \left( \frac{x}{2} + b \right)$ is a horizontal shift of $2 \sin \left( \frac{x}{2} \right)$.  Since the picture shows a peak at $\frac{\pi}{2}$, we need to shift $y=2 \sin \left( \frac{x}{2} \right)$ to the left by $\frac{\pi}{2}$ to achieve this.
The function $2 \sin \left( \frac{x}{2} \right)$ shifted $\frac{\pi}{2}$ to the left is
$$
2 \sin \left( \frac{x+\frac{\pi}{2}}{2} \right) = 2 \sin \left( \frac{x}{2}+\frac{\pi}{4} \right)
$$
and so $b=\frac{\pi}{4}$.
A: $\sin\left( \dfrac x 2 + b \right) = 1$ when $\dfrac x 2 + b = \dfrac \pi 2,$ and that implies $x=\pi-2b.$
Since it appears that the point where the curve reaches its maximum is half-way between $0$ and $\pi/2,$ we have $\dfrac\pi 2 = \pi - 2b,$ and from that we get $b=\dfrac \pi 4.$
A bit of an optical illusion seems to afflict this graph: The positioning of the $\text{“}0\text{''}$ a bit to the right of the zero point, presumably to avoid putting it right on top of that vertical axis, makes it look as if that maximum point may be closer to $0$ than to $\pi/2.$ But closer inspection dispells that.
