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The function is \( y = \frac{6}{x^2 – x} \). Which is a vertical asymptote of this function? The x-axis spans from below negative 5 to beyond 5, and the y-axis spans from below negative 40 to above 20. The x-axis has a scale of 5 in increments of 1, and the y-axis has a scale of 20 in increments of 5. The red rational function consists of two convex curves in the first and second quadrants, with a narrow, bounded region with single peak below the x-axis. The first-quadrant convex curve starts from positive infinity near the vertical line x = 1, decreases sharply, and then approaches the horizontal asymptote along the positive x-axis. The second quadrant convex curve starts from positive infinity near x = 0, decreases toward the negative x-axis, and extends horizontally. A single narrow peak appears in the fourth quadrant near the point (0.5, negative 24), where the function briefly rises before decreasing again. This localized fluctuation is confined to a narrow, bounded region in between x = 0 and x = 1. 

Which of the following represents a rational function with a vertical asymptote at \( x = 4 \)?

What is the horizontal asymptote of \[f(x) = \frac{3x + 1}{x + 2}?\]

What is the degree of the numerator in the rational function \( f(x) = \frac{3x^3 + x}{2x^2 – 1} \), and what does it imply about the end behavior?

What is the vertical asymptote of \( f(x) = \frac{1}{x – 1} + 3 \)? The x-axis spans from below negative 2 to above 2, and the y-axis spans from below negative 5 to above 10. The x-axis has a scale of 2 in increments of 0.5, and the y-axis has a scale of 5 in increments of 1. The convex curve is in the first quadrant, passing through the points (1.5, 5) and (3, 3.5). It starts from positive infinity above the vertical asymptote near x= 1. It decreases steeply before leveling off as it approaches the horizontal asymptote near y= 3. The concave curve spans the fourth, first and second quadrants, passing through a point slightly left of (0.75, 0) and (0, 2). It starts from negative infinity below the vertical asymptote near x= 1, increasing steeply, and then approaching the horizontal asymptote near y =3 in the second quadrant. 

Rewrite the following in the form \( q(x) + \frac{r(x)}{d(x)} \): \[\frac{x^3 + 2x^2 – 3}{x^2 – 1}\]

If \( f(x) = \frac{1}{x} \) is transformed to \( g(x) = \frac{1}{x} – 3 \), what happens to the graph?

What happens to the graph of \( f(x) = \frac{1}{x} \) when it is replaced with \( f(x) = \frac{1}{2x} \)? The x-axis spans from below negative 2 to above 2, and the y-axis spans from below negative 5 to above 5. The x-axis has a scale of 2 in increments of 0.5, and the y-axis has a scale of 5 in increments of 1. The convex curve spans the first quadrant, passing through the points (0.25, 2) and (1, 0.5). It starts from positive infinity near the vertical asymptote at x = 0 and decreasing toward the horizontal asymptote at y = 0. The concave curve is in the third quadrant, passing through the points (negative 1, negative 0.5) and (negative 0.25, negative 2). It approaches negative infinity as it nears the vertical asymptote at x = 0 and levels out toward the horizontal asymptote near y= 0 as x moves left. 

Simplify: \[\frac{4}{x^2 – 9} – \frac{2}{x + 3}\]

Add the rational expressions: \[\frac{3x}{x^2 + 5x} + \frac{2}{x^2 + 5x}\]