QUESTION IMAGE
Question
use the drawing tools to form the correct answer on the provided graph. graph the line that represents the equation $y = -\frac{3}{2}x + 1$.
Step1: Identify the y - intercept
The equation of the line is in slope - intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. For the equation \(y=-\frac{3}{2}x + 1\), the y - intercept \(b = 1\). This means the line crosses the y - axis at the point \((0,1)\). So we plot the point \((0,1)\) on the graph.
Step2: Use the slope to find another point
The slope \(m=-\frac{3}{2}\). The slope is defined as \(\frac{\text{change in }y}{\text{change in }x}\), or \(\frac{\Delta y}{\Delta x}\). A slope of \(-\frac{3}{2}\) means that for a change in \(x\) of \(2\) (we can choose a positive value for \(\Delta x\) for simplicity), the change in \(y\) is \(- 3\) (because \(\frac{\Delta y}{\Delta x}=-\frac{3}{2}\)). Starting from the point \((0,1)\), if we move \(2\) units to the right (increase \(x\) by \(2\)) to \(x = 0+2=2\), then we move \(3\) units down (decrease \(y\) by \(3\)) from \(y = 1\) to \(y=1 - 3=-2\). So we get the point \((2,-2)\). We can also use a negative \(\Delta x\): if we move \(2\) units to the left (decrease \(x\) by \(2\)) from \(x = 0\) to \(x=- 2\), then we move \(3\) units up (increase \(y\) by \(3\)) from \(y = 1\) to \(y = 1+3 = 4\), getting the point \((-2,4)\).
Step3: Draw the line
After plotting two points (for example, \((0,1)\) and \((2,-2)\) or \((0,1)\) and \((-2,4)\)), we use the line tool to draw a straight line passing through these two points. This line represents the equation \(y =-\frac{3}{2}x+1\).
(Note: Since this is a graph - drawing problem, the final answer is the line drawn through the appropriate points as described above. If we were to describe the key points: the y - intercept at \((0,1)\) and another point such as \((2,-2)\) or \((-2,4)\) and the line connecting them.)
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The line is drawn through the points \((0,1)\) and \((2,-2)\) (or \((0,1)\) and \((-2,4)\)) on the provided coordinate grid.