What is the relationship between the range and the equation of the asymptote

Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function. (They can also arise in other contexts, such as logarithms, but you'll almost certainly first encounter asymptotes in the context of rationals.)  I'll give you an example:

This is a rational function. More to the point, this is a fraction. Can you have a zero in the denominator of a fraction? No. So if I set the denominator of the above fraction equal to zero and solve, this will tell me the values that x cannot be:

x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or –1

So x cannot be 6 or –1, because then I'd be dividing by zero.

The domain is the set of all x-values that I'm allowed to use. The only values that could be disallowed are those that give me a zero in the denominator. So I'll set the denominator equal to zero and solve.

x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = –4 or x = 2

Since I can't have a zero in the denominator, then I can't have x = –4 or x = 2 in the domain. This tells me that the vertical asymptotes (which tell me where the graph can not go) will be at the values x = –4 or x = 2.

answer: the equation matches the regression

step-by-step explanation:

∠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.