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Use the image to answer the question. Derive the equation of the line in slope-intercept form. 1, -1 0,-4

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To derive the equation of the line in slope-intercept form \(y = mx + b\) using the points (1, -1) and (0, -4), we follow these steps:

  1. Identify the points:

    • Point 1: \((x_1, y_1) = (1, -1)\)
    • Point 2: \((x_2, y_2) = (0, -4)\)

  2. Calculate the slope \(m\) using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting the values from the points: \[ m = \frac{-4 - (-1)}{0 - 1} = \frac{-4 + 1}{-1} = \frac{-3}{-1} = 3 \]

  3. Use the slope and one of the points to find the y-intercept \(b\). We'll use the point (0, -4) since it already provides the y-intercept directly: \[ b = -4 \]

  4. Write the equation in slope-intercept form: Substitute \(m\) and \(b\) into the equation \(y = mx + b\): \[ y = 3x - 4 \]

Thus, the equation of the line in slope-intercept form is: \[ y = 3x - 4 \]

Use the image to answer the question. Derive the equation of the line in slope-intercept form. 0,1 1,-4

To derive the equation of the line in slope-intercept form \(y = mx + b\) using the points (0, 1) and (1, -4), follow these steps:

  1. Identify the points:

    • Point 1: \((x_1, y_1) = (0, 1)\)
    • Point 2: \((x_2, y_2) = (1, -4)\)

  2. Calculate the slope \(m\) using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting the values from the points: \[ m = \frac{-4 - 1}{1 - 0} = \frac{-5}{1} = -5 \]

  3. Use the slope and one of the points to find the y-intercept \(b\). Since point \((0, 1)\) already provides the \(y\)-intercept directly: \[ b = 1 \]

  4. Write the equation in slope-intercept form: Substitute \(m\) and \(b\) into the equation \(y = mx + b\): \[ y = -5x + 1 \]

Thus, the equation of the line in slope-intercept form is: \[ y = -5x + 1 \]

Graph the linear equation y=−3x+2 .(1 point) Responses Image with alt text: An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid upward line with arrows at both ends is plotted on the graph. The line passes through the plotted points left parenthesis 0 comma 3 right parenthesis and left parenthesis 1 comma 5 right parenthesis. Image with alt text: A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points. The coordinates of the plotted points are as follows: left parenthesis negative 2 comma 2 right parenthesis, and left parenthesis 0 comma 1 right parenthesis. Image with alt text: A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points. The coordinates of the plotted points are as follows: left parenthesis 0 comma 2 right parenthesis, and left parenthesis 1 comma negative 1 right parenthesis.

To graph the linear equation \( y = -3x + 2 \), we can identify key points and plot them on a coordinate plane.

Steps to Graph the Equation:

  1. Identify the y-intercept: The equation is in slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Here, \(b = 2\). This means that the line crosses the y-axis at \( (0, 2) \).

  2. Identify the slope: The slope \(m = -3\) means that for each unit you move to the right (increase in \(x\)), you move 3 units down (decrease in \(y\)).

  3. Calculate another point using the slope: Starting from the y-intercept (0, 2):

    • Move right 1 unit to \(x = 1\) and down 3 units to \(y = -1\). This gives us the point \( (1, -1) \).

  4. Plot the points:

    • Plot \((0, 2)\) on the graph.
    • Plot \((1, -1)\) on the graph.

  5. Draw the line: Connect these points with a straight line extending in both directions with arrows indicating that it continues.

Analyzing the Responses:

Based on the provided options describing the graphs:

  • Option 1: Points \((0, 3)\) and \((1, 5)\) do not represent the equation \(y = -3x + 2\).
  • Option 2: Points \((-2, 2)\) and \((0, 1)\) also do not match the equation.
  • Option 3: Points \((0, 2)\) and \((1, -1)\) correctly represent the points derived from \(y = -3x + 2\).

Conclusion:

The correct graph of the equation \(y = -3x + 2\) is represented by Option 3, which includes the points \((0, 2)\) and \((1, -1)\). Thus, this graph accurately reflects the linear equation provided.