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 set \( y = 0 \) and solve for \( x \):

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

To make it easier to work with, we can multiply the entire equation by \(-1\):

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

Next, we can use the quadratic formula to find the zeros, where \( a = 1 \), \( b = -16 \), and \( c = 39 \). The quadratic formula is given by:

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

Let's first calculate the discriminant \( b^2 - 4ac \):

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

Now we can 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 possible 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 quadratic function are:

- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)

In summary:

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