Which of the following points represents a valid solution to the constraints represented by the inequalities shown in the graph? The x-axis spans from negative 5 to 5, and the y-axis spans from below 0 to above 5. Both axes have a scale of 5 in increments of 1. A solid black inverted V-shaped function peak at (negative 1, 5). It opens downward with a y-intercept of (0, 4), an x-intercept of 4, 0), while enclosing a gray-shaded region below it. A dashed purple V-shaped function has a sharp vertex at (2, negative 1). It opens upward, with x-intercepts of (1, 0) and (3, 0), y-intercept of (0, 1), while enclosing a purple-shaded region above it. The two shaded areas overlap with intersection points (negative 2.5, 3.5) and (3.5, 0.5). The overlap forms a dark purple rectangular region at the center of the graph, spanning the first, second, and slight of the fourth quadrants.

Simplify the following: \[\frac{\frac{2x^2}{x + 3}}{\frac{x}{x – 3}}\]

What is the effect of the transformation \( f(x) = \frac{1}{x} \) to \( g(x) = \frac{1}{x} + 2 \)?

For the function \( f(x) = \frac{1}{x} \), replacing it with \( \frac{1}{x + 3} \) results in:

If \( f(x) = \vert x \vert \), what transformation is represented by \( g(x) = -\vert x – 2 \vert + 3 \)?

For the function \( f(x) = \frac{1}{x^2 – 4} \), identify the vertical asymptotes.

The function is \( y = – \frac{1}{2x+5} + 1 \). 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 0, and the y-axis spans from below negative 10 to just above 10. 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 convex curve is in the second quadrant, passing approximately through the points (negative 2.5, 10) and (negative 5, 1). The curve starts from positive infinity above the vertical asymptote near x= negative 2.5, then decreases steeply, approaching the horizontal asymptote near y = 1. The concave curve spans the third, second and first quadrants, passing through the approximate points (negative 2.5, negative 10) and (1, 1). It starts from negative infinity approaching the vertical asymptote near x= negative 2.5, then rises approaching the horizontal asymptote near y = 1. 

Which of the following rational functions have a vertical asymptote at \( x = 3 \)?

Identify the horizontal asymptote for the rational function \( f(x) = \frac{4x + 3}{2x + 5} \).

Solve \( x^2 + 4 = 0 \).