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.)
smaller zero: x = ___; greater zero: x = ____

Answer 1

To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to solve the equation \( -x^2 + 16x - 39 = 0 \).

We can use the quadratic formula, which is given by:

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

For our quadratic equation:

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

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

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

Now, using the quadratic formula:

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

Now, we can calculate the two possible values for \( x \):

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

Thus, the zeros of the quadratic function are \( x = 3 \) and \( x = 13 \).

Now we can identify which is the smaller zero and which is the greater zero:

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

So the answers are:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)