To determine the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can set the function equal to zero and solve for \( x \):
\[ -x^2 + 16x - 39 = 0 \]
To make it easier to solve, we can multiply the equation by -1:
\[ x^2 - 16x + 39 = 0 \]
Now we can use the quadratic formula to find the zeros, where \( a = 1 \), \( b = -16 \), and \( c = 39 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100 \]
Now substitute the values into the quadratic formula:
\[ x = \frac{16 \pm \sqrt{100}}{2} \]
Since \( \sqrt{100} = 10 \), we have:
\[ x = \frac{16 \pm 10}{2} \]
This leads to two possible solutions:
- \( x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \)
- \( x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \)
Thus, the zeros of the quadratic function are \( x = 3 \) and \( x = 13 \).
When listing them in order, the smaller zero is:
\[ \text{smaller zero: } x = 3 \]
and the greater zero is:
\[ \text{greater zero: } x = 13 \]
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