Question
What happens to the x-intercepts of the function y=x2−25
when the graph is reflected over the x-axis? Explain.(1 point)
Responses
The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.
The x -intercepts remain the same because the y -values get negated during a reflection over the x -axis.
This function does not have any x-intercepts, so no changes are necessary.
This function does not have any x -intercepts, so no changes are necessary.
The x-values and y-values switch.
The x -values and y -values switch.
The x-intercepts become negated.
when the graph is reflected over the x-axis? Explain.(1 point)
Responses
The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.
The x -intercepts remain the same because the y -values get negated during a reflection over the x -axis.
This function does not have any x-intercepts, so no changes are necessary.
This function does not have any x -intercepts, so no changes are necessary.
The x-values and y-values switch.
The x -values and y -values switch.
The x-intercepts become negated.
Answers
GPT-4o mini
To understand what happens to the x-intercepts of the function \( y = x^2 - 25 \) when the graph is reflected over the x-axis, let's first determine the x-intercepts of the original function.
The x-intercepts occur where \( y = 0 \):
\[
0 = x^2 - 25
\]
Solving for \( x \),
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
Thus, the x-intercepts are at the points \( (5, 0) \) and \( (-5, 0) \).
When the graph is reflected over the x-axis, the y-values of all points are negated, but the x-values remain unchanged. This means the points where the original function has x-intercepts will not change their x-coordinates, and since the y-value was already zero at the intercepts, it remains zero after reflection.
Therefore, the x-intercepts remain at the same x-coordinates:
- \( (5, 0) \) stays \( (5, 0) \)
- \( (-5, 0) \) stays \( (-5, 0) \)
The correct response is:
**The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.**
The x-intercepts occur where \( y = 0 \):
\[
0 = x^2 - 25
\]
Solving for \( x \),
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
Thus, the x-intercepts are at the points \( (5, 0) \) and \( (-5, 0) \).
When the graph is reflected over the x-axis, the y-values of all points are negated, but the x-values remain unchanged. This means the points where the original function has x-intercepts will not change their x-coordinates, and since the y-value was already zero at the intercepts, it remains zero after reflection.
Therefore, the x-intercepts remain at the same x-coordinates:
- \( (5, 0) \) stays \( (5, 0) \)
- \( (-5, 0) \) stays \( (-5, 0) \)
The correct response is:
**The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.**