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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A line in the slope-intercept form is written as $y=mx+b,$ where $m$ is the slope and $b$ is the $y$-intercept. We want to find the graph of $y=3x−1.$ Since $b=-1$ we know that it has its $y$-intercept at $(0,-1).$ Let's mark this point in a coordinate plane.

The function has a the slope $m=3.$ With help of the slope we can find one more point on the line. $m=ΔxΔy ⇒m=3=13 $ This means that the function moves $3$ units in the $y$-direction when it moves $1$ unit in the $x$-direction. Using this we can find another point on the graph.

We can now draw the graph of the function by connecting these points with a line

We can use the same technique to draw the graph of $y=-2x+4.$ Since $b=4$ the $y$-intercept is at $(0,4).$ The slope is $-2.$ Therefore, we can find another point on the graph by starting at the $y$-intercept and move $1$ unit to the right and $2$ units down. Let's mark these points and connect them with a line.

The last function we want to draw is $y=-x−3,$ which we can write as $y=-1x−3.$ We see that $m=-1$ and $b=-3.$ We draw the graph by first marking the $y$-intercept, $(0,-3).$ Using that $m=-1$ we find another point on the graph by moving $1$ unit to the right and $1$ unit down. We then connect the points with a line.