How can Kendra determine if the function is actually
1 answer:
You might be interested in
Answer:
Remember the property:
a^-1 = (1/a)^1
and:
(a/b)^n = (a^n)/(b^n)
A table for a function like:
Is just completed as:
So, here we have:
y = f(x) = (1/6)^x
To complete the table, we need to find:
f(-1)
and
f(2)
So let's find these two values:
f(-1) = (1/6)^-1 = (6/1)^1 = 6
and the other value is:
f(2) = (1/6)^2 = 1/36
Then the complete table is: