To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to set the function equal to zero:
\[ -x^2 + 16x - 39 = 0 \]
To solve this quadratic equation, we can rearrange it by multiplying through by -1 to make the leading coefficient positive:
\[ x^2 - 16x + 39 = 0 \]
Next, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 1 \), \( b = -16 \), and \( c = 39 \). Plugging these values into the formula gives:
- Calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 39 = 256 - 156 = 100 \]
- Now apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{100}}{2 \cdot 1} = \frac{16 \pm 10}{2} \]
This gives us two possible solutions:
\[ x = \frac{26}{2} = 13 \]
\[ x = \frac{6}{2} = 3 \]
Now, we have the zeros \( x = 13 \) and \( x = 3 \).
According to your request to list the smaller zero first, the zeros of the function are:
\[ \boxed{3} \quad \text{and} \quad \boxed{13} \]