Select the correct answer from each drop-down menu. Consider the function f(x) = 3x + 1 and the graph of the function g(x) shown below.

Function transformation involves changing the position of a function.

The graph of g(x) is the graph of f(x) translated 2 units right, and

The function is given as:

The graph of g(x) passes through (2,1) and (0,-5).

Start by calculating the slope (m)

So, we have:

The equation is then calculated as:

So, we have:

By comparing and

The graph of f(x) is shifted 2 units to the right

Read more about function transformation at:

link

Translation of 2 units to the right

g(x) = f(x-2)

Step-by-step explanation:

Given the original function f (x) = 3x +1, we can get the function plotted by translating the graph 2 units to the right.

As seen in the attached image, the original function f(x) is shown in orange, while the translation 2 units to the right is the function in blue. Notice that in the original function the crossing of the x-axis occurs at x = -1/3, and translating the function to the right 2 units moves the crossing two full units to the right, making it now cross the x-axis at x = 5/3

Such translation in 2 units to the right corresponds to a subtraction of 2 units from the x-variable in the original functional expression:

g (x) = f (x - 2) g (x) = 3(x - 2) + 1

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TYSM

∠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.

the answer is b.

step-by-step explanation: