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QUESTION IMAGE

use the drawing tools to form the correct answer on the provided graph.…

Question

use the drawing tools to form the correct answer on the provided graph. graph the line that represents the equation $y = -\frac{3}{2}x + 1$.

Explanation:

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.)

Answer:

The line is drawn through the points \((0,1)\) and \((2,-2)\) (or \((0,1)\) and \((-2,4)\)) on the provided coordinate grid.