To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can set \( y = 0 \) and solve for \( x \):
\[ -x^2 + 16x - 39 = 0 \]
To make it easier to work with, we can multiply the entire equation by \(-1\):
\[ x^2 - 16x + 39 = 0 \]
Next, we can use the quadratic formula to find the zeros, where \( a = 1 \), \( b = -16 \), and \( c = 39 \). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's first calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100 \]
Now we can substitute the values into the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{100}}{2 \cdot 1} = \frac{16 \pm 10}{2} \]
This gives us two possible solutions:
- \( x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \)
- \( x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \)
Thus, the zeros of the quadratic function are:
- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)
In summary:
smaller zero: \( x = 3 \); greater zero: \( x = 13 \)
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