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Rational functions often have vertical asymptotes, which occur when the function's denominator is zero. These asymptotes are vertical lines that the graph of the function approaches but never touches or crosses. To find them, set the denominator to zero and solve for the variable.

In our given function, \(y = \frac{2x^2 - 5x}{2x + 3}\), the denominator \(2x + 3\) becomes zero when \(x = -\frac{3}{2}\). Therefore, the vertical asymptote is at \(x = -\frac{3}{2}\). It's crucial to check whether the numerator is also zero at this value because if both the numerator and denominator are zero, it indicates a removable discontinuity, not an asymptote. However, in this case, the numerator remains non-zero, confirming a true vertical asymptote at \(x = -\frac{3}{2}\).

Remember, vertical asymptotes indicate a behavior change in the graph; near these values, the function may head towards positive or negative infinity.

To determine where a graph intersects the x-axis for a rational function, find the x-intercepts by setting the numerator equal to zero and solving for \(x\). The reason is that at these points, the value of the function is zero.

For our function, \(2x^2 - 5x = 0\). By factoring, we get \(x(2x - 5) = 0\), revealing x-intercepts at \(x = 0\) and \(x = \frac{5}{2}\).

These intercepts mean that the graph crosses the x-axis at these points, indicating that the function's output is zero here. You can visualize this as the places where the curve touches or crosses the horizontal x-axis, a common characteristic in polynomial or rational functions.

The y-intercept of a graph is the point where the graph crosses the y-axis. For a rational function, the y-intercept is found by setting \(x = 0\) since the y-axis is defined by \(x=0\).

In our function \(y = \frac{2x^2 - 5x}{2x + 3}\), substituting \(x = 0\) gives us:
\[ y = \frac{2(0)^2 - 5(0)}{2(0) + 3} = 0 \] Therefore, the y-intercept is \((0,0)\).

This point tells us that the function's graph touches the y-axis at the origin, meaning when \(x=0\), the output is zero. It's a key point easily identifiable on the graph and helpful in understanding the overall graph structure.

Local extrema of a function, such as maxima or minima, are points where a function reaches a local high or low. For rational functions, finding these points involves calculus, specifically finding the derivative and then determining where the derivative equals zero or is undefined (critical points).

For our function, we use the quotient rule for derivatives since it is a fraction of polynomials. Let \(u = 2x^2 - 5x\) and \(v = 2x + 3\). Their derivatives are \(u' = 4x - 5\) and \(v' = 2\). Applying the quotient rule:
\[ y' = \frac{(v u' - u v')}{v^2} \]
Evaluate where \(y' = 0\) to find the critical points. These points are potential locations for local extrema.

To conclude whether these points are maxima or minima, further analysis or a graphing utility might be employed. Any calculated local extrema provide vital insights into the function's shape, indicating peaks or valleys.