To determine which of the given quadratic functions has real zeros at [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex], we can use the fact that the roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are related to its factors.
When a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has roots [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex], it can be written in the factored form as:
[tex]\[ a (x - \alpha) (x - \beta) = 0 \][/tex]
Given the roots [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex], we can express the quadratic equation as:
[tex]\[ (x + 10)(x + 6) \][/tex]
To convert this factored form back to the standard quadratic form [tex]\( x^2 + bx + c \)[/tex], expand the factors:
[tex]\[ (x + 10)(x + 6) = x^2 + 6x + 10x + 60 \][/tex]
[tex]\[ = x^2 + 16x + 60 \][/tex]
So the quadratic function that has real zeros at [tex]\( x = -10 \)[/tex] and [tex]\( x = -6 \)[/tex] is:
[tex]\[ f(x) = x^2 + 16x + 60 \][/tex]
Comparing this with the given options, we can see that the correct choice is:
[tex]\[ f(x) = x^2 + 16x + 60 \][/tex]
Therefore, the correct answer is:
[tex]\[ f(x) = x^2 + 16x + 60 \][/tex]
