Which function is the result of vertically shrinking [tex]f(x)=(x+1)^2[/tex] by a factor of [tex]\frac{1}{6}[/tex]?

A) [tex]f(x)=\left(\frac{1}{6} x + 1\right)^2[/tex]
B) [tex]f(x)=6(x+1)^2[/tex]
C) [tex]f(x)=(6 x + 1)^2[/tex]
D) [tex]f(x)=\frac{1}{6} (x+1)^2[/tex]



Answer :

To determine the result of vertically shrinking the function [tex]\(f(x) = (x + 1)^2\)[/tex] by a factor of [tex]\(\frac{1}{6}\)[/tex], let's go through the steps carefully:

1. Original Function: The original function given is [tex]\(f(x) = (x + 1)^2\)[/tex].

2. Vertical Shrinking: Vertically shrinking a function by a factor of [tex]\(\frac{1}{6}\)[/tex] means that every output value of the function will be scaled by [tex]\(\frac{1}{6}\)[/tex]. This involves multiplying the entire function by [tex]\(\frac{1}{6}\)[/tex].

3. Applying the Vertical Shrink: We take the given function [tex]\(f(x) = (x + 1)^2\)[/tex] and multiply it by [tex]\(\frac{1}{6}\)[/tex]:
[tex]\[ f(x) = \frac{1}{6} \cdot (x + 1)^2 \][/tex]

4. Resulting Function: The transformed function after applying the vertical shrinking factor is:
[tex]\[ f(x) = \frac{1}{6}(x + 1)^2 \][/tex]

Now, we compare this with the given options:

A) [tex]\(f(x) = \left(\frac{1}{6}x + 1\right)^2\)[/tex] - This is incorrect because it involves shrinking the [tex]\(x\)[/tex]-term inside the squared terms, not applying a vertical shrink to the whole function.

B) [tex]\(f(x) = 6(x + 1)^2\)[/tex] - This is incorrect because it represents a vertical stretch, not a shrink.

C) [tex]\(f(x) = (6x + 1)^2\)[/tex] - This is incorrect because it changes the [tex]\(x\)[/tex]-term inside the squared terms, not applying a vertical shrink to the whole function.

D) [tex]\(f(x) = \frac{1}{6}(x + 1)^2\)[/tex] - This is correct because it correctly applies the vertical shrink by a factor of [tex]\(\frac{1}{6}\)[/tex].

Therefore, the answer is [tex]\(\boxed{D}\)[/tex].