Question

determined the zeros of the quadric function y=-x^2 + 16x - 39

Answer 1

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

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

Rearranging the equation gives:

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

Now we can apply the quadratic formula, which is given by:

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

Here, \( a = 1 \), \( b = -16 \), and \( c = 39 \). Plugging in these values:

1. Calculate the discriminant:

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

2. Now apply the quadratic formula:

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

3. This yields two potential solutions:

\[
x_1 = \frac{16 + 10}{2} = \frac{26}{2} = 13
\]

\[
x_2 = \frac{16 - 10}{2} = \frac{6}{2} = 3
\]

Thus, the zeros of the quadratic function \( y = -x^2 + 16x - 39 \) are \( x = 13 \) and \( x = 3 \).