How would you describe the relationship between the real zeros and the [tex]$x$[/tex]-intercepts of the function [tex]$y=\log_4(x-2)$[/tex]?

A. When you set the function equal to zero, the solution is [tex]$x=6$[/tex]; therefore, the graph has an [tex]$x$[/tex]-intercept at [tex]$x=6$[/tex].
B. When you substitute [tex]$x=0$[/tex] into the function, there is no solution; therefore, the graph will not have any [tex]$x$[/tex]-intercepts.
C. Since there is an asymptote at [tex]$x=2$[/tex], the graph will not have an [tex]$x$[/tex]-intercept; therefore, the function will have no real zeros.
D. When you set the function equal to zero, the solution is [tex]$x=3$[/tex]; therefore, the graph has an [tex]$x$[/tex]-intercept at [tex]$x=3$[/tex].



Answer :

To find the relationship between the real zeros and the [tex]\( x \)[/tex]-intercepts of the function [tex]\( y = \log_4(x - 2) \)[/tex], let's follow these steps:

1. Set the function equal to zero to find [tex]\( x \)[/tex]-intercepts:
[tex]\[ y = \log_4(x - 2) = 0 \][/tex]

2. Solve for [tex]\( x \)[/tex]:
The equation [tex]\( \log_4(x - 2) = 0 \)[/tex] can be rewritten in exponential form as:
[tex]\[ 4^0 = x - 2 \][/tex]
Since [tex]\( 4^0 = 1 \)[/tex]:
[tex]\[ 1 = x - 2 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 1 + 2 \][/tex]
[tex]\[ x = 3 \][/tex]

So, the solution to the equation [tex]\( \log_4(x - 2) = 0 \)[/tex] is [tex]\( x = 3 \)[/tex]. Therefore, the function has an [tex]\( x \)[/tex]-intercept at [tex]\( x = 3 \)[/tex].

3. Check for vertical asymptotes:
The function [tex]\( y = \log_4(x - 2) \)[/tex] has a vertical asymptote where the argument of the logarithm is zero. The argument is [tex]\( x - 2 \)[/tex], so we set it to zero:
[tex]\[ x - 2 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 2 \][/tex]

Thus, the function has a vertical asymptote at [tex]\( x = 2 \)[/tex].

From this analysis, we can conclude:

- The function [tex]\( y = \log_4(x - 2) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( x = 3 \)[/tex].
- There is a vertical asymptote at [tex]\( x = 2 \)[/tex].

Given the choices, the correct relationship is:

"When you set the function equal to zero, the solution is [tex]\( x = 3 \)[/tex]; therefore, the graph has an [tex]\( x \)[/tex]-intercept at [tex]\( x = 3 \)[/tex]."