Question

find g(x), where g(x) is the translation 5 units right of f(x) = |x|. write your answer in the form a|x - h| + k, where a, h, and k are integers. g(x) =

Explanation:

To determine the function \( g(x) \) which is the translation of \( f(x) = |x| \) 5 units to the right, we use the rules for horizontal translations of functions.

Step 1: Recall the transformation rule for horizontal shifts

For a function \( f(x) \), the transformation \( f(x - h) \) shifts the graph of \( f(x) \) \( h \) units to the right (if \( h>0 \)) and \( h \) units to the left (if \( h < 0 \)).

Step 2: Apply the transformation to \( f(x)=|x| \)

We want to shift \( f(x)=|x| \) 5 units to the right. Using the horizontal shift rule, if we let \( h = 5 \), then the new function \( g(x) \) is given by:

$$ g(x)=f(x - 5) $$

Since \( f(x)=|x| \), we substitute \( x-5 \) into \( f(x) \):

$$ g(x)=|x - 5| $$

Step 3: Write in the form \( a|x - h|+k \)

We compare \( |x - 5| \) with \( a|x - h|+k \). Here, \( a = 1 \) (the coefficient of the absolute value term), \( h=5 \) (the value inside the absolute value with the subtraction), and \( k = 0 \) (there is no vertical shift, so the constant term is 0).

So, \( g(x)=1|x - 5|+0 \)

\( g(x)=\boxed{1|x - 5|+0} \)

Answer:

To determine the function \( g(x) \) which is the translation of \( f(x) = |x| \) 5 units to the right, we use the rules for horizontal translations of functions.

Step 1: Recall the transformation rule for horizontal shifts

For a function \( f(x) \), the transformation \( f(x - h) \) shifts the graph of \( f(x) \) \( h \) units to the right (if \( h>0 \)) and \( h \) units to the left (if \( h < 0 \)).

Step 2: Apply the transformation to \( f(x)=|x| \)

We want to shift \( f(x)=|x| \) 5 units to the right. Using the horizontal shift rule, if we let \( h = 5 \), then the new function \( g(x) \) is given by:

$$ g(x)=f(x - 5) $$

Since \( f(x)=|x| \), we substitute \( x-5 \) into \( f(x) \):

$$ g(x)=|x - 5| $$

Step 3: Write in the form \( a|x - h|+k \)

We compare \( |x - 5| \) with \( a|x - h|+k \). Here, \( a = 1 \) (the coefficient of the absolute value term), \( h=5 \) (the value inside the absolute value with the subtraction), and \( k = 0 \) (there is no vertical shift, so the constant term is 0).

So, \( g(x)=1|x - 5|+0 \)

\( g(x)=\boxed{1|x - 5|+0} \)