1. **State the problem:** We need to find an equation for the graph that decreases sharply from the top-left, passes through the origin near $\left(\frac{3\pi}{4},0\right)$, and goes steeply downward near $\left(\frac{3\pi}{2},-6\right)$. The graph resembles a transformed tangent function. 2. **Recall the tangent function properties:** The basic tangent function $y=\tan(x)$ has vertical asymptotes at $x=\frac{\pi}{2}+k\pi$ and zeros at $x=k\pi$ for integers $k$. It increases from $-\infty$ to $+\infty$ between asymptotes. 3. **Analyze the given points:** The zero near $\frac{3\pi}{4}$ suggests a horizontal shift because $\tan(x)$ is zero at multiples of $\pi$, but here zero is at $x=\frac{3\pi}{4}$. The steep decrease suggests a negative coefficient to flip the graph. 4. **Form the general transformed tangent function:** $$y = A \tan(B(x - C)) + D$$ where $A$ is amplitude (vertical stretch and reflection), $B$ affects period, $C$ is horizontal shift, and $D$ is vertical shift. 5. **Determine horizontal shift $C$:** Since zero is at $x=\frac{3\pi}{4}$, and tangent zeros occur at $x=C + k\frac{\pi}{B}$, set $C=\frac{3\pi}{4}$. 6. **Determine period and $B$:** The tangent period is $\frac{\pi}{B}$. The vertical asymptote near $x=\frac{3\pi}{2}$ is $\frac{3\pi}{4}$ away from zero at $\frac{3\pi}{4}$, so period is $2 \times \frac{3\pi}{4} - \frac{3\pi}{4} = \frac{\pi}{2}$. Thus, $$\frac{\pi}{B} = \frac{\pi}{2} \implies B=2$$ 7. **Determine amplitude $A$ and vertical shift $D$:** The graph goes down to about $-6$ near $x=\frac{3\pi}{2}$. The basic tangent at $\frac{\pi}{2}$ tends to $+\infty$, so to get a steep decrease, $A$ should be negative. Assume $D=0$ (no vertical shift). Calculate $y$ at $x=\frac{3\pi}{2}$: $$y = A \tan\left(2\left(\frac{3\pi}{2} - \frac{3\pi}{4}\right)\right) = A \tan\left(2 \times \frac{3\pi}{4}\right) = A \tan\left(\frac{3\pi}{2}\right)$$ Since $\tan\left(\frac{3\pi}{2}\right)$ is undefined (vertical asymptote), the function tends to $-\infty$ if $A<0$, matching the graph. Check value near zero at $x=\frac{3\pi}{4}$: $$y = A \tan(0) = 0$$ which matches the zero. 8. **Estimate $A$ from the value near $x=\pi$:** For example, at $x=\pi$, $$y = A \tan\left(2\left(\pi - \frac{3\pi}{4}\right)\right) = A \tan\left(2 \times \frac{\pi}{4}\right) = A \tan\left(\frac{\pi}{2}\right)$$ Again undefined, so the steepness is consistent. Given the graph reaches about $-6$, choose $A = -6$ for the vertical stretch and reflection. 9. **Final equation:** $$y = -6 \tan\left(2\left(x - \frac{3\pi}{4}\right)\right)$$ This matches the graph's behavior: zero at $x=\frac{3\pi}{4}$, vertical asymptotes at $x=\frac{3\pi}{4} \pm \frac{\pi}{4}$, and steep decrease. **Answer:** $$y = -6 \tan\left(2\left(x - \frac{3\pi}{4}\right)\right)$$