To find 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 work with, let's multiply the entire equation by -1:
\[ x^2 - 16x + 39 = 0 \]
Next, we can use the quadratic formula to find the zeros of the equation. The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 1 \), \( b = -16 \), and \( c = 39 \). Plugging these values into the formula:
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 39 = 256 - 156 = 100 \]
- Now, 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 potential 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:
\[ x = 13 \quad \text{and} \quad x = 3 \]