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:
- \( x - 3 = \sqrt{2} \)
- \( x - 3 = -\sqrt{2} \)
Now, solving for \( x \) in both cases:
- \( x = 3 + \sqrt{2} \)
- \( x = 3 - \sqrt{2} \)
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).