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 given by:

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

where \( a = -1 \), \( b = 16 \), and \( c = -39 \).

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

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

Now we can apply the quadratic formula:

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

Calculating further:

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

This gives us two solutions:

1. For the positive case:
\[
x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3
\]

2. For the negative case:
\[
x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13
\]

The zeros of the function are \( x = 3 \) and \( x = 13 \).

To specify which is the smaller and which is the greater zero, we write:

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

Therefore,
- smaller zero: \( x = 3 \);
- greater zero: \( x = 13 \).