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

What is the y-intercept of \[f(x) = \frac{5x}{x + 2}?\] The x-axis spans from below negative 10 to just above 5, and the y-axis spans from just below negative 20 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 convex curve is in the second quadrant, passing through the points (negative 3, 15) and (negative 6, 7.5). The curve starts from positive infinity above the vertical asymptote near x= negative 2, then decreases, approaching the horizontal asymptote near y = 5. The concave curve spans the third and the first quadrants, passing through the approximate points (negative 1.5, negative 15) and (3, 2.5). It starts from negative infinity approaching the vertical asymptote near x= negative 2, then rises approaching the horizontal asymptote near y = 5. 

Rewrite the equation in standard form and identify the conic:   \(x^2 + y^2 – 12x – 4y + 7 = 0\)