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
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 become negated.
The x -intercepts become negated.
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-values and y-values switch.
when the graph is reflected over the x-axis? Explain.(1 point)
Responses
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 become negated.
The x -intercepts become negated.
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-values and y-values switch.
Answers
Answered by
GPT-4o mini
To determine 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 \( y = 0 \):
\[
x^2 - 25 = 0
\]
Factoring gives:
\[
(x - 5)(x + 5) = 0
\]
Thus, the x-intercepts are \( x = 5 \) and \( x = -5 \).
When we reflect the graph of a function \( y = f(x) \) over the x-axis, the new function becomes \( y = -f(x) \). Therefore, reflecting \( y = x^2 - 25 \) over the x-axis gives:
\[
y = -(x^2 - 25) = -x^2 + 25
\]
To find the new x-intercepts of the reflected function, we set \( y = 0 \) again:
\[
0 = -x^2 + 25
\]
This simplifies to:
\[
x^2 = 25
\]
Thus, the new x-intercepts are also \( x = 5 \) and \( x = -5 \).
The x-intercepts remain unchanged because they are defined as the values of \( x \) that make \( y = 0 \). Reflecting over the x-axis negates the \( y \)-values but does not affect the \( x \)-values where \( y \) is zero.
Thus, 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 \( y = 0 \):
\[
x^2 - 25 = 0
\]
Factoring gives:
\[
(x - 5)(x + 5) = 0
\]
Thus, the x-intercepts are \( x = 5 \) and \( x = -5 \).
When we reflect the graph of a function \( y = f(x) \) over the x-axis, the new function becomes \( y = -f(x) \). Therefore, reflecting \( y = x^2 - 25 \) over the x-axis gives:
\[
y = -(x^2 - 25) = -x^2 + 25
\]
To find the new x-intercepts of the reflected function, we set \( y = 0 \) again:
\[
0 = -x^2 + 25
\]
This simplifies to:
\[
x^2 = 25
\]
Thus, the new x-intercepts are also \( x = 5 \) and \( x = -5 \).
The x-intercepts remain unchanged because they are defined as the values of \( x \) that make \( y = 0 \). Reflecting over the x-axis negates the \( y \)-values but does not affect the \( x \)-values where \( y \) is zero.
Thus, the correct response is:
**The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.**