To determine which function has real zeros at [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex], we need to evaluate each of the given functions by finding their roots. The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are found using the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Let's examine each function:
1. Function [tex]\( f(x) = x^2 + 16x + 60 \)[/tex]:
[tex]\[
x = \frac{{-16 \pm \sqrt{{16^2 - 4 \cdot 1 \cdot 60}}}}{2 \cdot 1} = \frac{{-16 \pm \sqrt{{256 - 240}}}}{2} = \frac{{-16 \pm \sqrt{16}}}{2} = \frac{{-16 \pm 4}}{2}
\][/tex]
[tex]\[
x = \frac{{-16 + 4}}{2} = -6 \quad \text{or} \quad x = \frac{{-16 - 4}}{2} = -10
\][/tex]
The roots are [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex].
2. Function [tex]\( f(x) = x^2 - 16x + 60 \)[/tex]:
[tex]\[
x = \frac{{16 \pm \sqrt{{(-16)^2 - 4 \cdot 1 \cdot 60}}}}{2 \cdot 1} = \frac{{16 \pm \sqrt{{256 - 240}}}}{2} = \frac{{16 \pm \sqrt{16}}}{2} = \frac{{16 \pm 4}}{2}
\][/tex]
[tex]\[
x = \frac{{16 + 4}}{2} = 10 \quad \text{or} \quad x = \frac{{16 - 4}}{2} = 6
\][/tex]
The roots are [tex]\( x = 10 \)[/tex] and [tex]\( x = 6 \)[/tex].
3. Function [tex]\( f(x) = x^2 + 4x + 60 \)[/tex]:
[tex]\[
x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 1 \cdot 60}}}}{2 \cdot 1} = \frac{{-4 \pm \sqrt{{16 - 240}}}}{2} = \frac{{-4 \pm \sqrt{-224}}}{2}
\][/tex]
The discriminant is negative ([tex]\(16 - 240 = -224\)[/tex]), thus the equation has no real roots.
4. Function [tex]\( f(x) = x^2 - 4x + 60 \)[/tex]:
[tex]\[
x = \frac{{4 \pm \sqrt{{(-4)^2 - 4 \cdot 1 \cdot 60}}}}{2 \cdot 1} = \frac{{4 \pm \sqrt{{16 - 240}}}}{2} = \frac{{4 \pm \sqrt{-224}}}{2}
\][/tex]
The discriminant is negative ([tex]\(16 - 240 = -224\)[/tex]), thus the equation has no real roots.
Comparing the results, we see that the function [tex]\( f(x) = x^2 + 16x + 60 \)[/tex] is the one that has real zeros at [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex].
Therefore, the function with real zeros at [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex] is:
[tex]\[ f(x) = x^2 + 16x + 60 \][/tex]
