Question

Which functions could represent a reflection over the y-axis of the given function? Check all that apply.

g(x) = –1/2(4)x
g(x) = 0.5(4)–x
g(x) = 2(4)x
g(x) =1/2(1/4) x
g(x) = 1/2 (1/4)–x

Answer 1

corect answers are for sure B and D -just took the test !

Answer 2

g(x) = 0.5(4)–x and g(x) =1/2(1/4) x

Step-by-step explanation

To reflect a function over the y-axis we use the rule:

`f(-x)` reflects the graph of
`f(x)` over the y-axis. In other words, to reflect
`f(x)` over the y-axis, we just need to replace
`x` with
`-x`.

Let's find
`f(x)` first using the standard exponential growth function:
`f(x)=a(1+b)^x`

From our graph we can infer that
`f(x)=(1)/(2)` when
`x=0`, so let's replace those values in our function:

`f(x)=a(1+b)^x`

`(1)/(2) =a(1+b)^0`

`(1)/(2) =a(1)`

`(1)/(2) =a`

We can also infer that
`f(x)=8` when
`x=2`. Since we already know that
`a=(1)/(2)`, let's replace the values to find
`b` an complete our function:

`f(x)=a(1+b)^x`

`8=(1)/(2) (1+b)^2`

`16=(1+b)^2`

`√(16) =1+b`

`4=1+b`

`b=3`

So putting it all together:

`f(x)=a(1+b)^x`

`f(x)=(1)/(2) (1+3)^x`

`f(x)=(1)/(2) (4)^x`

Finally, we can apply the reflection rule to find the reflection of
`f(x)` over the y-axis:

`f(-x)=(1)/(2) (4)^(-x)`

`g(x)=(1)/(2) (4)^(-x)`

Since
`(1)/(2) =0.5`,
`g(x)=(1)/(2) (4)^(-x)` and
`g(x)=0.5(4)^(-x)` are equivalent.

You can also express
`g(x)=(1)/(2) (4)^(-x)` as
`g(x)=(1)/(2) ((1)/(4) )^x` using laws of exponents.