To determine the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to set \( y \) to zero and solve for \( x \):
\[ 0 = -x^2 + 16x - 39 \]
Rearranging the equation gives:
\[ x^2 - 16x + 39 = 0 \]
Now we can apply the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -16 \), and \( c = 39 \). Plugging in these values:
- Calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100 \]
- Now apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{100}}{2(1)} = \frac{16 \pm 10}{2} \]
- This yields two potential solutions:
\[ x_1 = \frac{16 + 10}{2} = \frac{26}{2} = 13 \]
\[ x_2 = \frac{16 - 10}{2} = \frac{6}{2} = 3 \]
Thus, the zeros of the quadratic function \( y = -x^2 + 16x - 39 \) are \( x = 13 \) and \( x = 3 \).