Solved on Feb 12, 2024

Graph the linear function $y = -2x$ .

STEP 1

Assumptions
1. The equation given represents a direct variation between $y$ and $x$ .
2. The constant of variation $k$ is $-2$ .
3. The graph of the direct variation is a straight line passing through the origin.
4. We need to plot at least two points to draw the line.

STEP 2

Understand the form of a direct variation equation, which is $y = kx$ , where $k$ is the constant of variation.

STEP 3

Identify the constant of variation $k$ in the given equation.
$k = -2$

STEP 4

Since the graph of a direct variation is a straight line through the origin, we know one point is $(0,0)$ .

STEP 5

Choose a value for $x$ to find a second point on the line. Let's choose $x = 1$ .

STEP 6

Substitute $x = 1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot 1$

STEP 7

Calculate the $y$ value for $x = 1$ .
$y = -2$

STEP 8

Now we have a second point on the line: $(1, -2)$ .

STEP 9

Choose another value for $x$ to find a third point, which can help confirm the linearity of the graph. Let's choose $x = -1$ .

STEP 10

Substitute $x = -1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot (-1)$

STEP 11

Calculate the $y$ value for $x = -1$ .
$y = 2$

STEP 12

Now we have a third point on the line: $(-1, 2)$ .

STEP 13

Plot the points $(0,0)$ , $(1, -2)$ , and $(-1, 2)$ on a coordinate plane.

STEP 14

Draw a straight line through the points to represent the graph of the direct variation.
The graph of the direct variation $y = -2x$ is a straight line passing through the origin and the points $(1, -2)$ and $(-1, 2)$ .

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Graph the linear function $y = -2x$ .

STEP 1

Assumptions
1. The equation given represents a direct variation between $y$ and $x$ .
2. The constant of variation $k$ is $-2$ .
3. The graph of the direct variation is a straight line passing through the origin.
4. We need to plot at least two points to draw the line.

STEP 2

Understand the form of a direct variation equation, which is $y = kx$ , where $k$ is the constant of variation.

STEP 3

Identify the constant of variation $k$ in the given equation.
$k = -2$

STEP 4

Since the graph of a direct variation is a straight line through the origin, we know one point is $(0,0)$ .

STEP 5

Choose a value for $x$ to find a second point on the line. Let's choose $x = 1$ .

STEP 6

Substitute $x = 1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot 1$

STEP 7

Calculate the $y$ value for $x = 1$ .
$y = -2$

STEP 8

Now we have a second point on the line: $(1, -2)$ .

STEP 9

Choose another value for $x$ to find a third point, which can help confirm the linearity of the graph. Let's choose $x = -1$ .

STEP 10

Substitute $x = -1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot (-1)$

STEP 11

Calculate the $y$ value for $x = -1$ .
$y = 2$

STEP 12

Now we have a third point on the line: $(-1, 2)$ .

STEP 13

Plot the points $(0,0)$ , $(1, -2)$ , and $(-1, 2)$ on a coordinate plane.

STEP 14

Draw a straight line through the points to represent the graph of the direct variation.
The graph of the direct variation $y = -2x$ is a straight line passing through the origin and the points $(1, -2)$ and $(-1, 2)$ .

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Graph the linear function y=−2xy = -2xy=−2x.

STEP 1

Assumptions1. The equation given represents a direct variation between y y y and x x x.2. The constant of variation k k k is −2 -2 −2.3. The graph of the direct variation is a straight line passing through the origin.4. We need to plot at least two points to draw the line.

STEP 2

Understand the form of a direct variation equation, which is y=kx y = kx y=kx, where k k k is the constant of variation.

STEP 3

Identify the constant of variation k k k in the given equation.k=−2 k = -2 k=−2

STEP 4

Since the graph of a direct variation is a straight line through the origin, we know one point is (0,0) (0,0) (0,0).

STEP 5

Choose a value for x x x to find a second point on the line. Let's choose x=1 x = 1 x=1.

STEP 6

Substitute x=1 x = 1 x=1 into the equation to find the corresponding y y y value.y=−2⋅1 y = -2 \cdot 1 y=−2⋅1

STEP 7

Calculate the y y y value for x=1 x = 1 x=1.y=−2 y = -2 y=−2

STEP 8

Now we have a second point on the line: (1,−2) (1, -2) (1,−2).

STEP 9

Choose another value for x x x to find a third point, which can help confirm the linearity of the graph. Let's choose x=−1 x = -1 x=−1.

STEP 10

Substitute x=−1 x = -1 x=−1 into the equation to find the corresponding y y y value.y=−2⋅(−1) y = -2 \cdot (-1) y=−2⋅(−1)

STEP 11

Calculate the y y y value for x=−1 x = -1 x=−1.y=2 y = 2 y=2

STEP 12

Now we have a third point on the line: (−1,2) (-1, 2) (−1,2).

STEP 13

Plot the points (0,0) (0,0) (0,0), (1,−2) (1, -2) (1,−2), and (−1,2) (-1, 2) (−1,2) on a coordinate plane.

STEP 14

Draw a straight line through the points to represent the graph of the direct variation.The graph of the direct variation y=−2x y = -2x y=−2x is a straight line passing through the origin and the points (1,−2) (1, -2) (1,−2) and (−1,2) (-1, 2) (−1,2).

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Graph the linear function $y = -2x$ .

Assumptions
1. The equation given represents a direct variation between $y$ and $x$ .
2. The constant of variation $k$ is $-2$ .
3. The graph of the direct variation is a straight line passing through the origin.
4. We need to plot at least two points to draw the line.

Understand the form of a direct variation equation, which is $y = kx$ , where $k$ is the constant of variation.

Identify the constant of variation $k$ in the given equation.
$k = -2$

Since the graph of a direct variation is a straight line through the origin, we know one point is $(0,0)$ .

Choose a value for $x$ to find a second point on the line. Let's choose $x = 1$ .

Substitute $x = 1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot 1$

Calculate the $y$ value for $x = 1$ .
$y = -2$

Now we have a second point on the line: $(1, -2)$ .

Choose another value for $x$ to find a third point, which can help confirm the linearity of the graph. Let's choose $x = -1$ .

Substitute $x = -1$ into the equation to find the corresponding $y$ value.
$y = -2 \cdot (-1)$

Calculate the $y$ value for $x = -1$ .
$y = 2$

Now we have a third point on the line: $(-1, 2)$ .

Plot the points $(0,0)$ , $(1, -2)$ , and $(-1, 2)$ on a coordinate plane.

Draw a straight line through the points to represent the graph of the direct variation.
The graph of the direct variation $y = -2x$ is a straight line passing through the origin and the points $(1, -2)$ and $(-1, 2)$ .