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 find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we can use the quadratic formula:

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

Here, the coefficients are:

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

Now, first, we need to calculate the discriminant \( b^2 - 4ac \):

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

Now we can calculate the roots using the quadratic formula:

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

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

1. Using the plus sign: \[ x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3 \]

2. Using the minus sign: \[ x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13 \]

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

Finally, when listing the zeros:

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

Therefore, the answer is:

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