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F(x)=-3log(2)(x+1)+2 on th he asymptote and any two points with int

User MkMan
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Final answer:

The asymptote of the function f(x)=-3log2(x+1)+2 is x=-1. Two points on this function with integer y-values include (1, -1) and (3, -4), found by substituting appropriate x-values that result in the logarithm argument being a power of 2.

Step-by-step explanation:

The student is asking about the function f(x)=-3log2(x+1)+2, particularly in terms of its asymptote and two points where the function yields integer values.

The asymptote of this logarithmic function is a vertical line where the function is undefined, and that occurs when the argument of the logarithm is zero. In this case, the argument is (x+1), meaning the asymptote is the vertical line x=-1.

To find two points with integer values, we set up the equation f(x)=-3log2(x+1)+2 = k, where k is an integer. Since this may not always give a clear solution for x, we can choose specific x values that make the argument of the logarithm powers of 2 as this will result in integers after applying the logarithm.

For example, if we choose x=1, the function simplifies to f(1)=-3log2(2)+2 = -3(1)+2 = -1, which gives us one point (1, -1). Similarly, for x=3, we get f(3)=-3log2(4)+2 = -3(2)+2 = -4, giving us the point (3, -4). Thus, the two points are (1, -1) and (3, -4).

User Robou
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