1. **State the problem:** We need to write an equation of the form $y = a \sin bx$ or $y = a \cos bx$ for a sinusoidal graph with amplitude 4, midline $y=0$, and period $\frac{4\pi}{3}$. The graph crosses the origin going upward, has a maximum at $\left(\frac{2\pi}{3},4\right)$, a minimum at $\left(2\pi,-4\right)$, and the next maximum at $\left(\frac{10\pi}{3},4\right)$.\n\n2. **Identify amplitude $a$:** Amplitude is the distance from the midline to a maximum or minimum. Given amplitude is 4, so $a=4$.\n\n3. **Find $b$ using the period formula:** The period $T$ of $y = a \sin bx$ or $y = a \cos bx$ is given by $$T = \frac{2\pi}{b}.$$\nGiven $T = \frac{4\pi}{3}$, solve for $b$: $$b = \frac{2\pi}{T} = \frac{2\pi}{\frac{4\pi}{3}} = \frac{2\pi \times 3}{4\pi} = \frac{6}{4} = \frac{3}{2}.$$\n\n4. **Determine whether to use sine or cosine:** The graph crosses the origin going upward, which matches the behavior of $y = a \sin bx$ since $\sin 0 = 0$ and the sine function increases from 0 at $x=0$.\n\n5. **Write the equation:** Using $a=4$ and $b=\frac{3}{2}$, the equation is $$y = 4 \sin \left( \frac{3}{2} x \right).$$\n\n6. **Verify with given points:**\n- At $x=0$, $y=4 \sin(0) = 0$, matches origin crossing upward.\n- At $x=\frac{2\pi}{3}$, $y=4 \sin \left( \frac{3}{2} \times \frac{2\pi}{3} \right) = 4 \sin(\pi) = 0$, but the problem states maximum 4 here, so check carefully.\n\nSince $y=4 \sin \left( \frac{3}{2} x \right)$ gives zero at $x=\frac{2\pi}{3}$, but the graph has a maximum there, the sine function is not correct.\n\nTry cosine: $y=4 \cos \left( \frac{3}{2} x \right)$.\n- At $x=0$, $y=4 \cos(0) = 4$, but the graph crosses origin at 0, so cosine is shifted.\n\nTry phase shift for sine: $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ or $y=4 \cos \left( \frac{3}{2} x \right)$ shifted.\n\nAlternatively, since the graph crosses origin going upward and has maximum at $x=\frac{2\pi}{3}$, the function is a sine shifted left by $\frac{\pi}{3}$.\n\nUse $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ or equivalently $y=4 \cos \left( \frac{3}{2} x \right)$.\n\nCheck $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ at $x=0$: $$4 \sin \left( 0 - \frac{\pi}{2} \right) = 4 \sin \left( -\frac{\pi}{2} \right) = 4 \times (-1) = -4,$$ which is not crossing origin upward.\n\nTry $y=4 \sin \left( \frac{3}{2} x + \frac{\pi}{2} \right)$ at $x=0$: $$4 \sin \left( \frac{\pi}{2} \right) = 4 \times 1 = 4,$$ not zero.\n\nTry $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{3} \right)$ at $x=\frac{2\pi}{3}$: $$4 \sin \left( \frac{3}{2} \times \frac{2\pi}{3} - \frac{\pi}{3} \right) = 4 \sin \left( \pi - \frac{\pi}{3} \right) = 4 \sin \left( \frac{2\pi}{3} \right) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.46,$$ close but not 4.\n\nTry $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{6} \right)$ at $x=\frac{2\pi}{3}$: $$4 \sin \left( \pi - \frac{\pi}{6} \right) = 4 \sin \left( \frac{5\pi}{6} \right) = 4 \times \frac{1}{2} = 2,$$ less than 4.\n\nTry $y=4 \sin \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ at $x=\frac{2\pi}{3}$: $$4 \sin \left( \pi - \frac{\pi}{2} \right) = 4 \sin \left( \frac{\pi}{2} \right) = 4,$$ perfect maximum.\n\nCheck at $x=0$: $$4 \sin \left( 0 - \frac{\pi}{2} \right) = 4 \sin \left( -\frac{\pi}{2} \right) = -4,$$ which is minimum, but the graph crosses origin going upward, so this is not matching.\n\nTry $y=4 \cos \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ at $x=0$: $$4 \cos \left( -\frac{\pi}{2} \right) = 4 \times 0 = 0,$$ crosses origin.\n\nDerivative at $x=0$ for $y=4 \cos \left( \frac{3}{2} x - \frac{\pi}{2} \right)$ is positive (going upward), so this matches the graph behavior.\n\nTherefore, the equation is $$y = 4 \cos \left( \frac{3}{2} x - \frac{\pi}{2} \right).$$