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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: \[\frac{3x}{x^2 + 5x} + \frac{2}{x^2 + 5x}\]

What are the vertical asymptotes of \[f(x) = \frac{x + 1}{x^2 – 2}?\] “The x-axis spans from below negative 4 to just above 4, and the y-axis spans from below negative 10 to just 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 leftmost branch is a sharp concave curve in the third quadrant, starting from negative infinity near x = negative 1.5, increasing steeply, and then abruptly approaching the horizontal asymptote near y = 0.  The middle branch is between the asymptotes, decreasing from positive infinity near x = negative 1.5 in the second quadrant, passing through the point (negative 1,0) in a flat pattern and continuing downward past negative infinity near x = 1.5 in the fourth quadrant.   The rightmost branch is a sharp convex curve in the first quadrant, starting from positive infinity near x= 1.5 and decreasing steeply before leveling off as it approaches the horizontal asymptote near y= 0.  “

How does the graph of \( f(x) = \frac{1}{x} \) change when replaced by \( 2f(x) \)?

Which polynomial could represent the graph? 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. 

Which of the following describes the graph of a polynomial with a root at \( x = 2 \) with multiplicity 2?

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).

What are the remaining roots of \( P(x) = x^3 + 2x^2 + 7x + 6 \) if one root is \( x = -1 \)?