Find the equation of the line parallel to the graph of \( 2 x+y=3 \) that contains the point \( (-4,2) \) Write the answer in slope-intercept form.

To find the equation of the line parallel to the graph of \( 2x + y = 3 \) that contains the point \( (-4, 2) \), we first need to determine the slope of the given line. 1. **Convert the equation to slope-intercept form (y = mx + b)**: \[ 2x + y = 3 \] Subtract \( 2x \) from both sides: \[ y = -2x + 3 \] From this, we can see that the slope \( m \) of the line is \( -2 \). 2. **Since parallel lines have the same slope**, the slope of the line we want to find is also \( -2 \). 3. **Use the point-slope form of the equation of a line**: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is the point on the line and \( m \) is the slope. Here, \( (x_1, y_1) = (-4, 2) \) and \( m = -2 \). Plugging in these values: \[ y - 2 = -2(x + 4) \] 4. **Simplify the equation**: Distributing the \( -2 \): \[ y - 2 = -2x - 8 \] Now, add \( 2 \) to both sides: \[ y = -2x - 6 \] Thus, the equation of the line parallel to \( 2x + y = 3 \) that contains the point \( (-4, 2) \) in slope-intercept form is: \[ \boxed{y = -2x - 6} \]