Use the graph of f(x) to explain the relationship between the real zeros of f(x) and its intercept(s).

f(x) has one real zero at –2 because the graph of the function has an intercept at (0, –2).
f(x) has two real zeros at –4 and –2 because the graph of the function has intercepts at (–4, 0) and (0, –2).
f(x) has no real zeros because the graph of the function does not pass through (0, 0).
f(x) has one real zero at –4 because the graph of the function has an intercept at (–4, 0).

f(x) has one real zero at –4 because the graph of the function has an intercept at (–4, 0).

Step-by-step explanation:

I got it correct for the test

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

dont think so

step-by-step explanation: