For the function \( f(x) = \frac{x}{x – 3} \), the vertical asymptote is at:

For the function \( f(x) = \frac{2}{x – 1} \), the vertical asymptote is at:

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

What type of discontinuity occurs at \( x = -1 \) for \( f(x) = \frac{x^2 – 1}{x + 1} \)?

How can you find the x-intercept of \[f(x) = \frac{3x + 6}{x – 4}?\]

Which is one of the transformations to  represented in the graph? The x-axis spans from negative 5 to 10, and the y-axis spans from below 0 to just above 20. The x-axis has a scale of 5 in increments of 1, and the y-axis has a scale of 10 in increments of 2. The blue parabolic function opens upward and has a minimum vertex at (3, 1). The parabola is symmetric around the vertical line passing through the vertex and passes through the coordinates (0, 20) and (6, 20).

For the function \[f(x) = \frac{x – 3}{x^2 – x – 6},\] what are the vertical asymptotes?

If \[f(x) = \frac{1}{x^2 – 4},\] what are the vertical asymptotes?

What is the degree of the polynomial shown? The x-axis spans from below 0 to 10, and the y-axis spans from negative 20 to just above 20. The x-axis has a scale of 5 in increments of 1 and the y-axis has a scale of 10 in increments of 2. The purple parabola opens upward, with its vertex at approximately (3.5, negative 19). The curve is symmetric around the vertical line passing through the vertex. It intersects the y-axis at (0, 20). It crosses the x-axis at (1, 0) and (6, 0), extending out of view at both ends. 

The function is \( y = \frac{8}{x -4} \). Which of the following transformations accurately describes a transformation which has been applied to the parent function \( y = \frac{1}{x} \)? The x-axis spans from below negative 5 to above 10, and the y-axis spans from below negative 20 to just 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 convex curve is in the first quadrant, passing through the points (5, 7.5) and (7, 2.5). It starts from positive infinity near the vertical asymptote at x = 4 and decreasing toward the horizontal asymptote at y = 0. The concave curve spans the third and fourth quadrants, passing through the points (1, negative 2.5) and (3.5, negative 15). It starts from negative infinity near the vertical asymptote at x = 4, crosses the y axis slightly above (0, negative 2.5) and levels out toward the horizontal asymptote near y = 0 as x moves left.