1. The problem asks to identify the graph of the function $$h(x) = -\tan\left(\frac{1}{2}x\right)$$ from two options. 2. The parent tangent function is $$\tan(x)$$, which has vertical asymptotes at $$x = \frac{\pi}{2} + k\pi$$ and period $$\pi$$. 3. The function $$h(x) = -\tan\left(\frac{1}{2}x\right)$$ involves two transformations: - Horizontal stretch by a factor of 2 because of the $$\frac{1}{2}$$ inside the argument, so the period becomes $$\frac{\pi}{\frac{1}{2}} = 2\pi$$. - Reflection about the x-axis due to the negative sign in front. 4. The period of $$h(x)$$ is $$2\pi$$, so vertical asymptotes occur at $$x = \pm \pi, \pm 3\pi, ...$$, which means the distance between vertical asymptotes is about 6.28 units. 5. The negative sign means the tangent branches decrease from left to right (opposite to the parent tangent function which increases). 6. Option A shows increasing tangent branches with vertical asymptotes spaced about 2 units apart, which does not match the period or reflection. 7. Option B shows decreasing tangent branches with vertical asymptotes spaced about 2 units apart, which matches the reflection but not the period. 8. However, the problem states the x-intercepts are spaced about 2 units apart in both graphs, which suggests the horizontal scale is compressed in the graphs. 9. Since the function has a horizontal stretch by 2, the period is doubled, so the distance between vertical asymptotes should be about 6.28 units, but the graphs show about 2 units, indicating the graphs are scaled differently. 10. The key difference is the direction of the branches: increasing for option A and decreasing for option B. 11. Because of the negative sign, the tangent branches of $$h(x)$$ decrease from left to right. 12. Therefore, the correct graph representing $$h(x) = -\tan\left(\frac{1}{2}x\right)$$ is option B. **Final answer:** Option B