PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE HELP ME PLEASE I REALLY REALLY NEED THIS DONE ASAP Based on the graph and information given, choose the correct answer below.

a.) f’ is not defined at x = 0, x = 1 and x = 6 because at these points the graph has cusps

b.) f’ is not defined at x = -2, x = 1 and x = 4 because at these points the graph has cusps

c.) f’ is not defined at x = 0, x = 1 and x = 4 because at these points the graph has corners

d.) f’ is not defined at x = -2, x = 1 and x = 4 because at these points the graph has corners

∠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.

boxplot

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).