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 become negated.
The x -intercepts become negated.

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.

1 answer

To determine the effect of reflecting the function \( y = x^2 - 25 \) over the x-axis, we need to analyze the concept of x-intercepts. The x-intercepts are found by setting \( y = 0 \):

\[ 0 = x^2 - 25 \] Solving for \( x \) gives:

\[ x^2 = 25 \ x = 5 \quad \text{or} \quad x = -5 \]

Thus, the x-intercepts of the function are at \( x = 5 \) and \( x = -5 \).

When we reflect the graph of a function over the x-axis, the new function becomes \( y = -f(x) \). For our function, this results in:

\[ y = -(x^2 - 25) = -x^2 + 25 \]

The x-intercepts of the new function are found by setting \( y = 0 \):

\[ 0 = -x^2 + 25 \ x^2 = 25 \ x = 5 \quad \text{or} \quad x = -5 \]

Thus, the x-intercepts remain the same at \( x = 5 \) and \( x = -5 \) after the reflection.

Therefore, the correct response is:

The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.