Graphing Linear Functions in Slope-Intercept Form

To determine the error(s), let's begin by analyzing the given equation. Then we can draw the correct graph.

Equation

Let's start by recalling the slope-intercept form of a line. $y={\color{#0000FF}{m}}x+{\color{#009600}{b}}$ In the above formula, ${\color{#0000FF}{m}}$ is the slope and ${\color{#009600}{b}}$ is the $y\text{-}$intercept. Let's now write the given equation. $y={\color{#0000FF}{\text{-} 3}}x+{\color{#009600}{2}}$ We can see that the slope is ${\color{#0000FF}{\text{-} 3}}.$ This means that to get from one point on the line to another, we should move $3$ units down and $1$ unit to the right. We can also see that the $y\text{-}$intercept is ${\color{#009600}{2}}.$ The graph should cross the $y\text{-}$axis at the point $(0,2).$

Graph

Now, let's take a look at the drawn graph.

In the graph we can see that Ron-Jon marked the $y\text{-}$intercept at $(0,\text{-}3).$ Additionally, to move from this to the $x$-intercept, we move ${\color{#0000FF}{1.5}}$ units to the right and ${\color{#009600}{3}}$ units up. Let's use these values to see what slope Ron-Jon's graph has. $m=\dfrac{\Delta y}{\Delta x}=\dfrac{{\color{#009600}{3}}}{{\color{#0000FF}{1.5}}}=2$ Thus, Ron-Jon's graph has the slope $m=2.$ Let's summarize this information and the one obtained from the equation.

We can see that the mistake Ron-Jon made when he tried to graph the function was that he switched the slope and the $y\text{-}$intercept. Let's graph the equation correctly. The $y\text{-}$intercept is at $(0,2).$ Using that the slope is $\text{-} 3$ we can find one more point on the graph. $m=\dfrac{{\color{#0000FF}{\Delta y}}}{{\color{#009600}{\Delta x}}} \quad \Rightarrow \quad m=\text{-} 3=\dfrac{{\color{#0000FF}{\text{-} 3}}}{{\color{#009600}{1}}}$ Thus, we can find another point on the graph $1$ unit to the right and $3$ units down.

We draw the graph by connecting these points with a line.