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. 

Subtract the following rational expressions: \[\frac{5x}{x + 2} – \frac{3x}{x + 2}\]

Rewrite the equation in standard form and identify the conic:   \(x^2 + y^2 + 6x + 8y + 9 = 0\)

The horizontal asymptote of \( f(x) = \frac{3x}{x + 2} \) is:

Multiply and simplify: \[\frac{x^2 – x}{x^2 + x – 2} \cdot \frac{x + 2}{x}\]

How do you determine the x-intercept of \[f(x) = \frac{x – 4}{2x + 1}?\]

When the graph of \( f(x) = \frac{1}{x} \) is reflected across the y-axis, what is the resulting function?

What is the effect of the negative sign in \( f(x) = -\frac{1}{x + 4} \)? The x-axis spans from below negative 6 to 2, and the y-axis spans from below negative 10 to above 10. The x-axis has a scale of 2 in increments of 0.5, and the y-axis has a scale of 10 in increments of 2. The function has a vertical asymptote at x= negative 4 and a horizontal asymptote at y = 0. The convex curve is in the second quadrant, passing through the points (negative 4.25, 4) and (negative 5, 1). The curve starts from positive infinity above the vertical asymptote near x= negative 4, then decreases sharply approaching the horizontal asymptote near y = 0. The concave curve spans the third and the fourth quadrants, passing through the approximate points (negative 3.75, negative 4) and (negative 3, negative 1). It starts from negative infinity approaching the vertical asymptote near x= negative 2, then rises steeply approaching the horizontal asymptote near y = 0.