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Graphing Linear Functions in Slope-Intercept Form

Graphing Linear Functions in Slope-Intercept Form 1.6 - Solution

arrow_back Return to 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=mx+b y={\color{#0000FF}{m}}x+{\color{#009600}{b}} In the above formula, m{\color{#0000FF}{m}} is the slope and b{\color{#009600}{b}} is the y-y\text{-}intercept. Let's now write the given equation. y=-3x+2 y={\color{#0000FF}{\text{-} 3}}x+{\color{#009600}{2}} We can see that the slope is -3.{\color{#0000FF}{\text{-} 3}}. This means that to get from one point on the line to another, we should move 33 units down and 11 unit to the right. We can also see that the y-y\text{-}intercept is 2.{\color{#009600}{2}}. The graph should cross the y-y\text{-}axis at the point (0,2).(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-y\text{-}intercept at (0,-3).(0,\text{-}3). Additionally, to move from this to the xx-intercept, we move 1.5{\color{#0000FF}{1.5}} units to the right and 3{\color{#009600}{3}} units up. Let's use these values to see what slope Ron-Jon's graph has. m=ΔyΔx=31.5=2 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.m=2. Let's summarize this information and the one obtained from the equation.

Slope yy-intercept
Equation -3\text{-} 3 22
Ron-Jon's graph 22 -3\text{-} 3

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-y\text{-}intercept. Let's graph the equation correctly. The y-y\text{-}intercept is at (0,2).(0,2). Using that the slope is -3\text{-} 3 we can find one more point on the graph. m=ΔyΔxm=-3=-31 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 11 unit to the right and 33 units down.

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