Question

determine the zeros of y = -x^2 + 16x - 39

Answer 1

To determine the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we set the equation equal to zero and solve for \( x \):

\[
-x^2 + 16x - 39 = 0
\]

We can multiply the entire equation by -1 to simplify it:

\[
x^2 - 16x + 39 = 0
\]

Now, we will use the quadratic formula to find the zeros:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, the coefficients are:
- \( a = 1 \)
- \( b = -16 \)
- \( c = 39 \)

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100
\]

Now we can find the zeros using the quadratic formula:

\[
x = \frac{-(-16) \pm \sqrt{100}}{2 \cdot 1} = \frac{16 \pm 10}{2}
\]

This gives us two solutions:

1. \( x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \)
2. \( x = \frac{16 - 10}{2} = \frac{6}{2} = 3 \)

Thus, the zeros of the function \( y = -x^2 + 16x - 39 \) are:

\[
x = 13 \quad \text{and} \quad x = 3
\]