The zeros of a quadratic function are the x-values at which the function intersects the x-axis. For the quadratic function [tex]\( y = -\frac{1}{2}x^2 + x + 4 \)[/tex], we determine the zeros by solving the equation [tex]\( -\frac{1}{2}x^2 + x + 4 = 0 \)[/tex].
Here is the step-by-step process to find the zeros:
1. Identify the coefficients: In the given quadratic equation [tex]\( y = -\frac{1}{2}x^2 + x + 4 \)[/tex], we have:
- [tex]\( a = -\frac{1}{2} \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = 4 \)[/tex]
2. Set up the quadratic formula:
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
3. Calculate the discriminant:
The discriminant ([tex]\( \Delta \)[/tex]) is found using:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
\Delta = 1^2 - 4(-\frac{1}{2})(4)
\][/tex]
[tex]\[
\Delta = 1 + 2 \cdot 4
\][/tex]
[tex]\[
\Delta = 1 + 8
\][/tex]
[tex]\[
\Delta = 9
\][/tex]
4. Find the square root of the discriminant:
[tex]\[
\sqrt{\Delta} = \sqrt{9} = 3
\][/tex]
5. Determine the two potential solutions:
Using the quadratic formula:
[tex]\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} \quad \text{and} \quad x_2 = \frac{-b - \sqrt{\Delta}}{2a}
\][/tex]
Substituting [tex]\( b = 1 \)[/tex], [tex]\( \sqrt{\Delta} = 3 \)[/tex], and [tex]\( a = -\frac{1}{2} \)[/tex]:
[tex]\[
x_1 = \frac{-1 + 3}{2(-\frac{1}{2})} = \frac{2}{-1} = -2
\][/tex]
[tex]\[
x_2 = \frac{-1 - 3}{2(-\frac{1}{2})} = \frac{-4}{-1} = 4
\][/tex]
Therefore, the zeros of the function [tex]\( y = -\frac{1}{2}x^2 + x + 4 \)[/tex] are [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex]. Thus, the correct answer is:
-2 and 4
