Question

Which graph is the result of reflecting f(x) = One-fourth(8)x across the y-axis and then across the x-axis?

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and incresaes into quadrant 1. It goes through the y-axis at (0, 0.25) and goes through (1, 2).
On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases in quadrant 2. It crosses the y-axis at (0, 0.25) and goes through (negative 1, 2).
On a coordinate plane, an exponential function approaches y = 0 in quadrant 3 and decreases into quadrant 4. It crosses the y-axis at (0, negative 0.25) and goes through (1, negative 2).
On a coordinate plane, an exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (negative 1, negative 2) and crosses the y-axis at (0, negative 0.25).

Answer 1

Answer:

On a coordinate plane, an exponential function increases in

quadrant 3 into quadrant 4 and approaches y = 0. It goes through

(negative 1, negative 2) and crosses the y-axis at (0, negative 0.25) ⇒ last answer

Step-by-step explanation:

* Lets explain how to solve the problem

- The function $f(x)=\frac{1}{4}(8)^{x}$ is reflected across the

  y-axis and then across the x- axis

- Lets revise the reflection of a function across the axes

- If the function f(x) reflected across the x-axis, then the new  function

 h(x) = - f(x)

- If the function f(x) reflected across the y-axis, then the new  function

 g(x) = f(-x)

∵ $f(x)=\frac{1}{4}(8)^{x}$ is reflected across the y-axis

- Change the sign of x

∴ Its image is g(x) where $g(x)=\frac{1}{4}(8)^{-x}$

∵ $g(x)=\frac{1}{4}(8)^{-x}$ is reflected across the x-axis

- Change the sign of y

∴ Its image is h(x) where $h(x)=-\frac{1}{4}(8)^{-x}$

∴ h(x) is the image of f(x) after reflected across the y-axis then

  reflected across the x-axis

* Look to the attached graph for more understand

- f(x) represented by red

- g(x) represented by blue

- h(x) represented by green

* From the graph

- The green graph is in the 3rd and 4th quadrants

- Approaches y = 0 (x-axis)

- point (-1 , -2) lies on it

- It cross the y-axis at point (0 , -0.25)

∴ The answer is the last one

On a coordinate plane, an exponential function increases in

quadrant 3 into quadrant 4 and approaches y = 0. It goes through

(negative 1, negative 2) and crosses the y-axis at (0, negative 0.25)

Answer 2

The correct answer is option d)

Step-by-step explanation:

Given :

$f(x) = \dfrac{1}{4}(8)^x$

Solution :

Reflecting f(x) across the y - axis,

Change the sign of x in f(x).

Its image is g(x), where

$g(x)= \dfrac{1}{4}(8)^-^x$

g(x) reflecting across the x-axis,

Change the sign of y in g(x).

Its image is h(x), where

$h(x)=-\dfrac{1}{4}(8)^-^x$

See the given graph, where

f(x) represented by red

g(x) represented by green

h(x) represented by purple

We can observe that exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (-1,-2) and crosses the y-axis at (0,-0.25).

Therefore the correct answer is option d) On a coordinate plane, an exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (-1,-2) and crosses the y-axis at (0,-0.25).

For more information, refer the link given below

https://brainly.com/question/17267403?referrer=searchResults