What happens to the x-intercepts of the function y=x^2−25 when the graph is reflected over the x-axis? Explain.(1 point)

Question

Responses

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

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.
The x -intercepts become negated.

This function does not have any x-intercepts, so no changes are necessary.

Answer 1

To understand what happens to the x-intercepts of the function \( y = x^2 - 25 \) when reflected over the x-axis, let's first identify the x-intercepts of the original function.

The x-intercepts are found by setting \( y = 0 \):

\[ 0 = x^2 - 25 \]

This can be factored as:

\[ 0 = (x - 5)(x + 5) \]

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

Now, reflecting the graph of the function over the x-axis involves negating the y-values. The new equation will be:

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

To find the new x-intercepts, we again set \( y = 0 \):

\[ 0 = -x^2 + 25 \]

This can be rewritten as:

\[ x^2 = 25 \]

Taking the square root gives:

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

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

The correct response is:

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