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We have been given the equation $y=3x−y=41 $ and asked to graph a line to match the function. Let's first identify our $y$-intercept, the value for $y$ when $x=0:$ $y=3⋅0−41 ⇒y=-41 $ Now we know that $(0,-41 )$ is part of our graphed line. Next, we should consider $3$ from the equation. What does $3$ tell us? It tells us that with each new $x$ value, the line increases by $3$ units. This is the slope of the line. We can also think of it like this: $m=13 =ΔxΔy , $ every time we move by $1$ unit to the right, the line increases by $3$ units. Let's graph the line.

b

Unlike Part A, there is no way to find the $y$-intercept for this line. We can choose for our function to start anywhere along the $y$-axis. What we do know is that, the line will decrease by $3$ units for every $9$ units we move to the right. Our slope can be written as: $ΔxΔy =9-3 ⇒-31 =m $ Let's assume that the $y$-intercept is $0$ and graph our line.

c

This function, $g(x)=x+3.5,$ will be graphed in a very similar way to the equation in Part A. This is because, although it is written with function notation, we can still find the $y$-intercept and the slope. Recall that the $y$-intercept it the value of the function when $x=0.$ Here we have: $g(0)=0+3.5⇒g(0)=3.5. $ From the function, we can also see that for each new value of $x,$ there is only an increase by $x.$ In other words: $m=1⇒11 =ΔxΔy , $ every time we move $1$ unit to the right, our function increases by $1$ unit. Let's graph this line.

d

We are not given information that would help us find the $y-$intercept for this line. We can, however, use the given $Δx$ and $Δy$ to find the slope of the line.
$slope=ΔxΔy =4−45−1 =04 $
But wait! We can't divide by zero, that's just not possible. This means that our line is **not** a function. There is no change in $x,$ so we have a vertical line. Let's see what this looks like in a coordinate plane.