214k views
2 votes
F(x)=-log_(4)(x-2)-3 on the axes asymptote and any two points with integer

1 Answer

2 votes

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
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories