To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can set the equation equal to zero:
\[ -x^2 + 16x - 39 = 0 \]
To make it easier to solve, we can multiply the entire equation by -1:
\[ x^2 - 16x + 39 = 0 \]
Now, we can use the quadratic formula to find the zeros. The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 1 \), \( b = -16 \), and \( c = 39 \).
Now we can calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100 \]
Now that we have the discriminant, we can apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{100}}{2(1)} = \frac{16 \pm 10}{2} \]
This gives us two possible solutions:
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For the plus sign: \[ x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \]
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For the minus sign: \[ x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \]
Thus, the zeros of the function \( y = -x^2 + 16x - 39 \) are:
\[ x = 13 \quad \text{and} \quad x = 3 \]
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