Graph of f(x)=log_2(x) has been transformed to g(x)=alog_2(x)+k. The transformed image passes through the points (1/4,-9) and (16,-6). Determine a and k.
No idea. Thanks!
well, did you not just plug in the points given?
g(1/4) = -9, so
-2a+k = -9
g(16) = -6, so
4a+k = -6
a = 1/2
k = -8
g(x) = 1/2 log_2(x) - 8
To determine the values of "a" and "k" in the transformation of the graph, we need to use the given points (1/4, -9) and (16, -6).
Let's start by finding the original equation of the graph, f(x) = log2(x).
1. (1/4, -9) point:
When we substitute the x-coordinate (1/4) into the original equation, we get:
-9 = log2(1/4)
To solve for this logarithmic equation, we can rewrite it using the exponent rule:
2^(-9) = 1/4
Simplifying further, we get:
1/512 = 1/4
Since this is not true, the given point (1/4, -9) does not lie on the original graph.
2. (16, -6) point:
When we substitute the x-coordinate (16) into the original equation, we get:
-6 = log2(16)
Using the logarithmic property, we can rewrite it as an exponent:
2^(-6) = 16
Simplifying further, we get:
1/64 = 16
Again, this is not true, so the given point (16, -6) does not lie on the original graph either.
Since neither of the points lies on the original graph, it means there may be some mistake or data inconsistency in the problem statement. Please double-check the given points or the original equation for accuracy.
If you have any other questions or need further assistance, feel free to ask!