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Review the graph of g(x).

On a coordinate plane, a curve goes through (negative 4, 8), curves down through (negative 2, 0), and approaches x = negative 1 in quadrant 3. Another curve approaches x = negative 1 in quadrant 2, has vertex (0, 0), and curves up through (2, 8). Another curve approaches x = 3 in quadrant 4 and curves up through (4, 16).

What are the one-sided limits of g(x) around the vertical asymptote x = –1?

Limit of g (x) as x approaches negative 1 minus = infinity and limit of g (x) as x approaches negative 1 plus = infinity
Limit of g (x) as x approaches negative 1 minus = infinity and limit of g (x) as x approaches negative 1 plus = negative infinity
Limit of g (x) as x approaches negative 1 minus = negative infinity and limit of g (x) as x approaches negative 1 plus = infinity
Limit of g (x) as x approaches negative 1 minus = negative infinity and limit of g (x) as x approaches negative 1 plus = negative infinity

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Answer:

The answer is C

Explanation:

I just got done with the unit test review

User Abbas Mulani
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