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F(x)=-log_(4)(x-2)-3 on the axes asymptote and any two points with integer

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Final answer:

The function f(x) = -log4(x-2) - 3 has a vertical asymptote at x = 2 and shifts down by 3 units. Two points with integer coordinates on this function are (6, -4) and (18, -5).

Step-by-step explanation:

The function given is f(x) = -log4(x-2) - 3. The inverse function of exponentiation is the logarithm, and the logarithmic function here is based on the base 4. The graph of this function will have a vertical asymptote at x = 2 because the logarithm becomes undefined at that point. The graph shifts 3 units down due to the '-3' at the end of the function. To find two points with integer coordinates, we can plug in values for x that result in the argument of the logarithm being a power of 4 and then subtract 3.

For example, if x = 6, then we have:

f(6) = -log4(6-2) - 3 = -log4(4) - 3 = -1 - 3 = -4

So one point is (6, -4). Next, if x = 18, then:

f(18) = -log4(18-2) - 3 = -log4(16) - 3 = -2 - 3 = -5

So another point is (18, -5). Thus, we have two points with integer coordinates: (6, -4) and (18, -5).

User Chris Woolum
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