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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-values and y-values switch.
The x -values and y -values switch.
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-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.
The x-intercepts become negated.
Responses
The x-values and y-values switch.
The x -values and y -values switch.
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-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.
The x-intercepts become negated.
Answers
There are no human answers yet.
Answered by
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, we first need to identify the x-intercepts of the original function.
The x-intercepts occur where \( y = 0 \):
\[
0 = x^2 - 25
\]
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
So, the x-intercepts of the function \( y = x^2 - 25 \) are at \( x = 5 \) and \( x = -5 \).
When the graph is reflected over the x-axis, the new function would be \( y = -(x^2 - 25) = -x^2 + 25 \).
In this new function, the x-intercepts are found by setting \( y = 0 \) again:
\[
0 = -x^2 + 25
\]
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
Thus, the x-intercepts remain at \( x = 5 \) and \( x = -5 \).
The correct response to the question 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 \( y = 0 \):
\[
0 = x^2 - 25
\]
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
So, the x-intercepts of the function \( y = x^2 - 25 \) are at \( x = 5 \) and \( x = -5 \).
When the graph is reflected over the x-axis, the new function would be \( y = -(x^2 - 25) = -x^2 + 25 \).
In this new function, the x-intercepts are found by setting \( y = 0 \) again:
\[
0 = -x^2 + 25
\]
\[
x^2 = 25
\]
\[
x = 5 \quad \text{or} \quad x = -5
\]
Thus, the x-intercepts remain at \( x = 5 \) and \( x = -5 \).
The correct response to the question is:
**The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.**
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