What happens to the x-intercepts of the function y=x2−25

Question

when the graph is reflected over the x-axis? Explain
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-intercepts become negated.
The x -intercepts become negated.

The x-values and y-values switch.
The x -values and y -values switch.

Answer 1

To analyze what happens to the x-intercepts of the function \( y = x^2 - 25 \) when the graph is reflected over the x-axis, we first determine the x-intercepts of the original function.

The x-intercepts occur where \( y = 0 \):

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

Thus, the x-intercepts of the function \( y = x^2 - 25 \) are at \( (5, 0) \) and \( (-5, 0) \).

Next, reflecting the graph over the x-axis modifies the function by negating the y-values. The new function after this reflection would be:

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

However, the x-intercepts depend only on where \( y = 0 \). Let's set the new function to zero to find the x-intercepts of the reflected function:

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

This shows that the x-intercepts remain unchanged at \( (5, 0) \) and \( (-5, 0) \) after the reflection.

So, the correct statement is: The x-intercepts remain the same because the y-values get negated during a reflection over the x-axis.