Question

Determine the zeros of the quadratic function y=−x2+16x−39 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x=; greater zero: x=

Answer 1

To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can use the quadratic formula:

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

For the given quadratic function, the coefficients are:

- \( a = -1 \)
- \( b = 16 \)
- \( c = -39 \)

First, we will calculate the discriminant:

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

Next, we apply the quadratic formula:

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

Now we will calculate both solutions:

1. For the positive root:

\[
x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3
\]

2. For the negative root:

\[
x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13
\]

So, the two zeros of the function are \( x = 3 \) and \( x = 13 \).

Now we list them in the required format:

- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)

Thus, the final answer is:

- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)