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In the graph of a rational function, stationary points are where the derivative is equal to zero. These points indicate where the graph has a horizontal tangent, essentially marking the spots where the slope changes sign, such as turning from positive to negative or vice versa. To find these points, take the derivative of the function and set it equal to zero.
For the function given, \( f(x) = \frac{4}{x^2} - \frac{2}{x} + 3 \), the first derivative is calculated as \( f'(x) = -\frac{8}{x^3} + \frac{2}{x^2} \). Setting \( f'(x) = 0 \) allows us to solve \(-\frac{8}{x^3} + \frac{2}{x^2} = 0\). Solving gives \( x = 4 \). This stationary point is a location where the function doesn't increase or decrease momentarily, producing a flat spot on the graph.
By substituting back into the original function, we find \( f(4) = \frac{11}{4} \). Therefore, the stationary point is at \( (4, \frac{11}{4}) \). Look for these points as they represent local maxima or minima in the graph.

Vertical and horizontal asymptotes are like boundaries for the graph of a rational function.
**Vertical Asymptotes** appear where the function becomes undefined, often occurring where the denominator is zero. In the function \( f(x) = \frac{4}{x^2} - \frac{2}{x} + 3 \), the denominator becomes zero when \( x = 0 \). Thus, there is a vertical asymptote at \( x = 0 \), shown as a dashed vertical line on the graph.
**Horizontal Asymptotes** suggest the behavior of the function as \( x \) approaches positive or negative infinity. It is determined by comparing the degrees of the numerator and the denominator. For this exercise, the function behaves with a horizontal asymptote at \( y = 0 \) because the numerator's degree (0, since it is a constant) is less than that of the denominator.
These asymptotes help visualize the limits within which the graph exists, guiding how the graph infinitely stretches without crossing the line of the asymptote.

Inflection points on a graph refer to the spots where the curvature changes direction. These are found by taking the second derivative of the function and setting it to zero. For the function provided, the second derivative is \( f''(x) = \frac{24}{x^4} - \frac{4}{x^3} \). Setting \( f''(x) = 0 \) leads to solving \( \frac{24}{x^4} - \frac{4}{x^3} = 0 \), yielding \( x = 6 \).
At \( x = 6 \), substituting into the original function gives \( f(6) = \frac{26}{9} \). The inflection point at \( (6, \frac{26}{9}) \) marks where the graph changes from being concave upward to concave downward or vice versa.
Finding these points is crucial for understanding the full shape of the graph. They reveal the soft transitions between different curvatures, helping to provide a more detailed picture of how the function behaves beyond just the link between high and low areas.