To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to solve the equation \( -x^2 + 16x - 39 = 0 \).
We can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our quadratic equation:
- \( a = -1 \)
- \( b = 16 \)
- \( c = -39 \)
First, we will calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 16^2 - 4(-1)(-39) = 256 - 156 = 100 \]
Now, using the quadratic formula:
\[ x = \frac{-16 \pm \sqrt{100}}{2(-1)} \] \[ x = \frac{-16 \pm 10}{-2} \]
Now, we can calculate the two possible values for \( x \):
- \( x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3 \)
- \( x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13 \)
Thus, the zeros of the quadratic function are \( x = 3 \) and \( x = 13 \).
Now we can identify which is the smaller zero and which is the greater zero:
- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)
So the answers are:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
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