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

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when the graph is reflected over the x-axis? Explain.(1 point)
Responses

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.
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 become negated.

1 answer

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 need to determine the x-intercepts of the function.

The x-intercepts occur when \( y = 0 \). Setting the equation to zero gives us:

\[ 0 = x^2 - 25 \]

Solving for \( x \) gives:

\[ x^2 = 25 \] \[ x = \pm 5 \]

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

When the graph is reflected over the x-axis, the y-values of the function change signs. Specifically, if we reflect the function \( y = x^2 - 25 \) over the x-axis, we get the new function:

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

Now we can observe that the x-intercepts do not depend on the y-values, but rather on where the function intersects the x-axis. The x-intercepts remain at \( x = 5 \) and \( x = -5 \) because the values of \( x \) that make \( y = 0 \) are unchanged.

Therefore, the correct response is:

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