In order to find the zeros of the function [tex]\( f(x) = x^2 + x - 12 \)[/tex], we need to solve the equation [tex]\( x^2 + x - 12 = 0 \)[/tex]. A zero of the function is a value of [tex]\( x \)[/tex] that makes the function equal to zero. This translates to finding the solutions [tex]\( x \)[/tex] for the quadratic equation.
Let's review the quadratic equation:
[tex]\[ x^2 + x - 12 = 0 \][/tex]
We can use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients from the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
For the given equation [tex]\( x^2 + x - 12 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 1 \)[/tex]
- [tex]\( c = -12 \)[/tex]
We plug these values into the quadratic formula:
[tex]\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{49}}{2} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{2} \][/tex]
Now, consider the two scenarios for the plus and minus in the equation:
1. [tex]\( x = \frac{-1 + 7}{2} = \frac{6}{2} = 3 \)[/tex]
2. [tex]\( x = \frac{-1 - 7}{2} = \frac{-8}{2} = -4 \)[/tex]
Therefore, the solutions to the equation are:
[tex]\[ x = 3 \][/tex]
[tex]\[ x = -4 \][/tex]
So, the zeros of the function [tex]\( f(x) = x^2 + x - 12 \)[/tex] are [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex].
The correct answer is:
C. [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex]
