Let's carefully analyze the behavior of the function [tex]\( y = -\frac{1}{2}x + 2 \)[/tex] around [tex]\( x = 4 \)[/tex] and consider the given intervals.
First, we need to understand what the function does at [tex]\( x = 4 \)[/tex].
1. Calculate the value of [tex]\( y \)[/tex] when [tex]\( x = 4 \)[/tex]:
[tex]\[
y = -\frac{1}{2}(4) + 2
\][/tex]
Simplify the expression:
[tex]\[
y = -\frac{1}{2} \cdot 4 + 2 = -2 + 2 = 0
\][/tex]
So, when [tex]\( x = 4 \)[/tex], the value of [tex]\( y \)[/tex] is 0.
2. Verify the behavior of the function on either side of [tex]\( x = 4 \)[/tex]:
- For [tex]\( x < 4 \)[/tex]:
[tex]\[
x < 4 \implies -\frac{1}{2}x + 2 > 0
\][/tex]
The function is positive over the interval [tex]\((- \infty, 4)\)[/tex].
- For [tex]\( x > 4 \)[/tex]:
[tex]\[
x > 4 \implies -\frac{1}{2}x + 2 < 0
\][/tex]
The function is negative over the interval [tex]\((4, \infty)\)[/tex].
Given that the function [tex]\( y \)[/tex] switches from positive to negative as [tex]\( x \)[/tex] moves through 4, the value of the function at [tex]\( x = 4 \)[/tex] is 0. This behavior is characteristic of what we call a zero or root of the function.
Therefore, the correct statement is:
C. There is a zero at [tex]\( x=4 \)[/tex] because the function is changing from positive to negative at [tex]\( x=4 \)[/tex].
