Determine the zeros of the quadratic function y=−x^2+16x−39

Question

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

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

First, we can rearrange it into standard form by multiplying the entire equation by -1:

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

Now, we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -16 \), and \( c = 39 \).

Calculating the discriminant \( b^2 - 4ac \):

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

Now substituting back into the quadratic formula:

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

Calculating the two potential values for \( x \):

1. For the positive case:

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

2. For the negative case:

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

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

In conclusion:

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

So, the final answer is:

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