Question

Find and interpret the zeros of y = -2(x-3)^2 + 4

Answers

GPT-4o mini
To find the zeros of the function \( y = -2(x-3)^2 + 4 \), we need to set \( y \) to zero and solve for \( x \):

\[
0 = -2(x-3)^2 + 4
\]

First, we can isolate the squared term:

\[
-2(x-3)^2 = -4
\]

Dividing both sides by -2 gives:

\[
(x-3)^2 = 2
\]

Next, we take the square root of both sides:

\[
x - 3 = \pm \sqrt{2}
\]

This results in two solutions:

1. \( x - 3 = \sqrt{2} \)
2. \( x - 3 = -\sqrt{2} \)

Now, solving for \( x \) in both cases:

1. \( x = 3 + \sqrt{2} \)
2. \( x = 3 - \sqrt{2} \)

Thus, the zeros of the function are:

\[
x = 3 + \sqrt{2} \quad \text{and} \quad x = 3 - \sqrt{2}
\]

### Interpretation

The zeros of the function represent the x-values where the graph of the function intersects the x-axis.

- The function given is a downward-opening parabola (due to the negative coefficient of the squared term).
- The roots \( x = 3 + \sqrt{2} \) and \( x = 3 - \sqrt{2} \) indicate that there are two points where the function equals zero.

In numerical terms:

- \( 3 + \sqrt{2} \) is approximately \( 4.414 \)
- \( 3 - \sqrt{2} \) is approximately \( 1.586 \)

These points indicate the locations on the x-axis where the function takes the value of zero, defining the bounds of the interval over which the function is positive (between these zeros) and negative (outside this interval).

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