To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can use the quadratic formula given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = -1 \), \( b = 16 \), and \( c = -39 \).
First, we need to calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 = 16^2 = 256 \] \[ 4ac = 4 \cdot (-1) \cdot (-39) = 156 \] \[ b^2 - 4ac = 256 - 156 = 100 \]
Now we can apply the quadratic formula:
\[ x = \frac{-16 \pm \sqrt{100}}{2 \cdot -1} \]
Calculating further:
\[ \sqrt{100} = 10 \] \[ x = \frac{-16 \pm 10}{-2} \]
This gives us two solutions:
-
For the positive case: \[ x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3 \]
-
For the negative case: \[ x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13 \]
The zeros of the function are \( x = 3 \) and \( x = 13 \).
To specify which is the smaller and which is the greater zero, we write:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
Therefore,
- smaller zero: \( x = 3 \);
- greater zero: \( x = 13 \).
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