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-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.

Answer 1

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

The x-intercepts occur where \( y = 0 \). For the function \( y = x^2 - 25 \), we set it equal to zero:

\[
x^2 - 25 = 0
\]

Factoring the equation, we have:

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

This gives us the solutions:

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

So, the x-intercepts are at the points \( (5, 0) \) and \( (-5, 0) \).

Now, when we reflect the graph over the x-axis, the y-values of all points on the graph become negated while their x-values remain unchanged. Thus, the new function after reflection can be expressed as:

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

However, this transformation affects the y-values, not the x-values. The x-intercepts are still determined by where the function equals zero. So we need to find the x-intercepts of the new function:

Setting \( -x^2 + 25 = 0 \):

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

This also gives us:

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

Thus, the x-intercepts remain the same despite the reflection over the x-axis.

Therefore, the correct response is:

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