Question

Determine the zeros of the quadratic function y = -x^2 + 16x - 39. (When listing the smaller zero, remember that negative values are always smaller than positive values.)

Answer 1

To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to set the function equal to zero:

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

To solve this quadratic equation, we can rearrange it by multiplying through by -1 to make the leading coefficient positive:

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

Next, we can use the quadratic formula:

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

In our equation, \( a = 1 \), \( b = -16 \), and \( c = 39 \). Plugging these values into the formula gives:

1. Calculate the discriminant:

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

2. Now apply the quadratic formula:

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

This gives us two possible solutions:

\[
x = \frac{26}{2} = 13
\]

\[
x = \frac{6}{2} = 3
\]

Now, we have the zeros \( x = 13 \) and \( x = 3 \).

According to your request to list the smaller zero first, the zeros of the function are:

\[
\boxed{3} \quad \text{and} \quad \boxed{13}
\]