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A function of direct variation fulfills the following form:

[tex] y = kx
[/tex]

Where,

x: independent variable

y: dependent variable

k: proportionality constant (k> 0)

Therefore, the value of k is given by:

[tex] k = \frac{y2-y1}{x2-x1}
[/tex]

Substituting values:

[tex] k = \frac{12.5-5}{5-2} [/tex]

The value of k is given by:

[tex] k = 2.5 [/tex] (k>0)

Answer:

The table that represents a direct variation function is:

option 4 (from left to right)

Direct variation function is the relationship between the two variables.

Form the given table only table 4 gives the same value as for this constant which is 2.5. Thus the table 4 represents a direct variation function.

What is  direct variation function?

Direct variation function is the relationship between the two variables. This relationship can be expressed by equation of data table, in which one variable varies with the other variable.

Given information-

The values of x and y given in the table.

The equation of direct variation function can be given as,

[tex]y=mx[/tex]

Here, [tex]m[/tex] is the constant.

The value of this constant can be given as,

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

For the direct variation function the value of this constants should be equal for the each value of [tex]x,y[/tex] in the table

.

Form the given table only table 4 gives the same value as for this constant as,

[tex]m=\dfrac{-2.5-(-7.5)}{-1-(-3)}=2.5\\m=\dfrac{5-(-2.5)}{2-(-1)}=2.5\\m=\dfrac{12.5-(5)}{5-(2)}=2.5\\m=\dfrac{25-(12.5)}{10-(5)}=2.5[/tex]

Thus form the given table only table 4 gives the same value as for this constant which is 2.5. Thus the table 4 represents a direct variation function.

Learn more about the direct variation function here;

https://brainly.com/question/17093618

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