Which graph represents a function with direct variation? A coordinate plane with a line passing through (negative 4, 0) and (0, negative 2). A coordinate plane with a line passing through (negative 5, 4) and (0, 3). A coordinate plane with a line passing through (negative 4, negative 6) and (0, 3). A coordinate plane with a line passing through (negative 1, negative 4), (0, 0) and (1, 4).

Respuesta :

Answer:

A line passing through the points (-1,-4),(0,0) and (1,4)

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form [tex]y/x=k[/tex] or [tex]y=kx[/tex]

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Verify each case

Part 1) A line passing through the points (-4,0) and (0,-2)

This line not represent a direct variation, because the line not passes through the origin.

Part 2) A line passing through the points (-5,4) and (0,3)

This line not represent a direct variation, because the line not passes through the origin.

Part 3) A line passing through the points (-4,-6) and (0,3)

This line not represent a direct variation, because the line not passes through the origin.

Part 4) A line passing through the points (-1,-4),(0,0) and (1,4)

The line passes through the origin

Find out the value of k

[tex]k=y/x[/tex]

For the point (-1,-4)

substitute

[tex]k=-4/-1=4[/tex]

For the point (1,4)

substitute

[tex]k=4/1=4[/tex]

The linear equation is  [tex]y=4x[/tex]

This line represent a direct variation

The graph of a function with a direct variation is a straight line that goes

through the origin.

The graph that represents a function with direct variation is; A coordinate

plane with a line passing through (-1, -4), (0, 0), and (1, 4)

Reason:

Required: To find the graph that represents a function with direct variation.

Solution:

A direct variation is a variation between two variables, x, and y, which can

be expressed in the form; y = k·x

From the above equation, by comparison with the equation of a straight

line, y = m·x + c,  we have that the graph of a function with a direct variation

has a slope of k, and a y-intercept of 0.

The x-intercept is also 0, given that at the x-intercept, y = 0, therefore;

0 = k·x

[tex]x = \dfrac{0}{k} = 0[/tex]

Therefore, the point (0, 0) is a point on the graph with a direct variation

From the given options, the option that has the point (0, 0) as a point on the

graph of the function is the option;

  • A coordinate plane with a line passing through (-1, -4), (0, 0), and (1, 4)

By verification, we have;

Slope of line [tex]\dfrac{0 - (-4)}{0 - (-1)} = 4[/tex]

Equation of line is y - (-4) = 4·(x - (-1)) = 4·x + 4

y = 4·x + 4 + (-4) = 4·x

y = 4·x

The equation of the graph is in the form y = k·x, where; k = 4.

Therefore;

The graph that represents a function with direct variation is; A

coordinate plane with a line passing through (-1, -4), (0, 0), and (1, 4)

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