Question

What transformation has changed the parent function f(x) = (.5)x to its new appearance shown in the graph below?

Answer 1

@ranga

Answer 2

f(x) = (0.5)^x
At x = 0, f(x) = 1.
But in the graph you see f(x) =1 when x = -2
So the graph has been shifted to the LEFT by 2 units.
That will happen when x is replaced by x + 2.
Therefore, the function shown in the graph is f(x+2)

Answer 3

Express 34 = x as a logarithmic equation.

log34 = x

log43 = x

log3x = 4

log4x = 3

The function f(x) = 15(2)x represents the growth of a frog population every year in a remote swamp. Elizabeth wants to manipulate the formula to an equivalent form that calculates every half-year, not every year. Which function is correct for Elizabeth's purposes?

f(x) = 15(2 ^1/2)2x

f(x) = 15(2)^x/2

f(x) = 15/2(2)^x

f(x) = 30(2)x

Answer 4

\[34~~or~~3^4?\]

Answer 5

3^4 I apoligize

Answer 6

\[
3^4 = x\\
\text{Take logarithm to the base 3 on both sides.}\\
4 = log_3(x)
\]

Answer 7

^^^ The third answer choice.

Answer 8

\[f(x) = 15(2)^x?\]

Answer 9

f(x)=15(2)^x/2

Answer 10

I am asking you if the original function is: f(x) = 15 * (2)^x where x is in years?

Answer 11

Yes sir it was.

Answer 12

Another one of those vague questions where they don't tell you what x represents in the modified formula.

Without any modifications, to calculate every half year all one has to do is put x = 1/2 in the original formula and calculate f(x). But that is not what they want.

Originally x was in years and now they have modified the meaning of x but they don't tell you what it is in the question. If we make the assumption that after modification, x = 1 represent 6 months, x = 2 represents 1 year, x = 3 represents 1.5 years, etc. then f(x) = 15 * 2^(x/2).

Answer 13

@ranga If Denise wanted to create a function that modeled an exponential function with base of 12 and what exponents were needed to reach specific values, how would she set up her function?

f(x) = x^12

f(x) = log12^x

f(x) = 12^x

f(x) = logx12

Answer 14

@ganeshie8

Answer 15

\[2^{2y} = 5\\\\
\text{Take logarithm to the base 2 on both sides}\\
2ylog_2(2) = log_2(5)\\
2y = log_2(5)\\
y = \frac 12log_2(5)\\\\
\text{Read the value of }log_2(5) \text{ from the graph and substitute above.}
\]

Answer 16

\[f(x) = 12^x\]