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When dealing with the behavior of a graph as the variable \( x \) approaches infinity, we look to understand how the function behaves at extreme values of \( x \). This concept helps us determine if the graph levels out, known as approaching a horizontal asymptote.
In the given function \( y = \frac{3}{2} \left( \frac{x}{x-1} \right)^{2/3} \), as \( x \rightarrow \pm \infty \), the fraction \( \frac{x}{x-1} \) tends to 1. This is because as \( x \) becomes very large, the -1 in the denominator becomes insignificant compared to the value of \( x \).
Thus, \( \left(\frac{x}{x-1}\right)^{2/3} \) approaches 1, and the overall function approaches \( y = \frac{3}{2} \times 1 = \frac{3}{2} \). This constant value suggests the presence of a horizontal asymptote at \( y = \frac{3}{2} \). This behavior means the graph will never cross this line at extreme values and remains stable around it.

Vertical asymptotes occur where a function becomes undefined, often leading the graph to shoot upwards or downwards to infinity. For our function, \( y = \frac{3}{2} \left( \frac{x}{x-1} \right)^{2/3} \), this happens when the denominator of the fraction \( \frac{x}{x-1} \) equals zero.
Solving \( x - 1 = 0 \), we find a vertical asymptote at \( x = 1 \). As \( x \) approaches 1 from the right (\( x \rightarrow 1^+ \)), \( \frac{x}{x-1} \) becomes very large, resulting in a sharp increase in \( y \), heading towards infinity.
From the left (\( x \rightarrow 1^- \)), \( \frac{x}{x-1} \) becomes very negative but its absolute value is still large. Since the component is inside a power of \( \frac{2}{3} \), \( y \) becomes a large positive number as well. Thus, the graph spikes upwards on either side toward infinity, confirming the vertical asymptote at \( x = 1 \).

Function discontinuities occur where a function is not defined, which typically show up in the graph as gaps, jumps, or asymptotic behavior. In our function, \( y = \frac{3}{2} \left( \frac{x}{x-1} \right)^{2/3} \), the discontinuity is at \( x = 1 \). This is where the denominator goes to zero, causing the expression inside the power to be undefined.
The function itself cannot produce a real value at \( x = 1 \), making it a point of discontinuity. In practical terms, you won't see a continuous line passing through this point on the graph.
Thus, we have to understand that around \( x=1 \), the function will show sharp spikes rather than a continuous curve due to this discontinuity. Other areas of the function like near \( x = -1 \), however, remain continuous as there is no cause for any break there, meaning the graph flows smoothly through these points.