Question

Determine the zeros of the quadratic function y=−x2+16x−39

Answer 1

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

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

To make it easier to solve, we can multiply the entire equation by -1:

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

Now, we can use the quadratic formula to find the zeros. The quadratic formula is given by:

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

In our equation, \( a = 1 \), \( b = -16 \), and \( c = 39 \).

Now we can calculate the discriminant \( b^2 - 4ac \):

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

Now that we have the discriminant, we can apply the quadratic formula:

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

This gives us two possible solutions:

1. For the plus sign:
\[
x = \frac{16 + 10}{2} = \frac{26}{2} = 13
\]

2. For the minus sign:
\[
x = \frac{16 - 10}{2} = \frac{6}{2} = 3
\]

Thus, the zeros of the function \( y = -x^2 + 16x - 39 \) are:

\[
x = 13 \quad \text{and} \quad x = 3
\]