Question

Please help. I don’t understand this math problem . I’ll mark you as brainliest if correct

Answer 1

Answer:

  $f(x)=3\log_4{(2-x)}$

Step-by-step explanation:

You know that a log function has a vertical asymptote at x=0 and an x-intercept at (1, 0). It trends upward to the right.

The given graph has a vertical asymptote at x=2, indicating the transformation involves a right-shift of 2 units. It trends upward to the left of the vertical asymptote, suggesting a reflection is involved. At 1 unit left of the vertical asymptote, the graph crosses y=0, so there is no horizontal scaling involved.

At this point, you can write a version of the function f(x) that incorporates the reflection and translation:

  f(x) = k·log(-(x -2)) = k·log(2-x)

To find the vertical scale factor k, we make use of the additional point that is given.

  f(-2) = 3

  k·log(2 -(-2)) = 3 . . . . substitute into our f(x) so far

  k = 3/log(4) . . . . . . . . solve for k

Now, the function can be written as ...

  $f(x)=\dfrac{3}{\log{4}}\log{(2-x)}=3\dfrac{\log{(2-x)}}{\log{4}}$

Using the "change of base" formula for logarithms we can compact this to ...

  $\boxed{f(x)=3\log_4{(2-x)}}$