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Study Guides > Intermediate Algebra

Graph Linear Functions Using Slope and y-Intercept

Learning Outcome

  • Graph a linear function using the slope and y-intercept
Another way to graph a linear function is by using its slope and y-intercept. Let us consider the following function.

f(x)=12x+1f\left(x\right)=\dfrac{1}{2}x+1

The function is in slope-intercept form, so the slope is 12\dfrac{1}{2}. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when x=0x=0. The graph crosses the y-axis at (0,1)(0, 1). Now we know the slope and the y-intercept. We can begin graphing by plotting the point (0,1)(0, 1). We know that the slope is rise over run, m=riserunm=\dfrac{\text{rise}}{\text{run}}. From our example, we have m=12m=\dfrac{1}{2}, which means that the rise is 11 and the run is 22. So starting from our y-intercept (0,1)(0, 1), we can rise 11 and then run 22, or run 22 and then rise 11. We repeat until we have a few points and then we draw a line through the points as shown in the graph below. graph of the line y = (1/2)x +1 showing the

A General Note: Graphical Interpretation of a Linear Function

In the equation f(x)=mx+bf\left(x\right)=mx+b
  • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis.
  • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

m=change in output (rise)change in input (run)=ΔyΔx=y2y1x2x1m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{\Delta y}{\Delta x}=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}

All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.)

How To: Given the equation for a linear function, graph the function using the y-intercept and slope

  1. Evaluate the function at an input value of zero to find the y-intercept.
  2. Identify the slope.
  3. Plot the point represented by the y-intercept.
  4. Use riserun\dfrac{\text{rise}}{\text{run}} to determine at least two more points on the line.
  5. Sketch the line that passes through the points.

Example

Graph f(x)=23x+5f\left(x\right)=-\dfrac{2}{3}x+5 using the y-intercept and slope.

Answer: Evaluate the function at x=0x=0 to find the y-intercept. The output value when x=0x=0 is 55, so the graph will cross the y-axis at (0,5)(0, 5). According to the equation for the function, the slope of the line is 23-\dfrac{2}{3}. This tells us that for each vertical decrease in the "rise" of 2–2 units, the "run" increases by 33 units in the horizontal direction. We can now graph the function by first plotting the y-intercept in the graph below. From the initial value (0,5)(0, 5), we move down 22 units and to the right 33 units. We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points. graph of the line y = (-2/3)x + 5 showing the change of -2 in y and change of 3 in x. The graph slants downward from left to right, which means it has a negative slope as expected.

Try It

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In the following video we show another example of how to graph a linear function given the y-intercepts and the slope. https://youtu.be/N6lEPh11gk8 In the last example, we will show how to graph another linear function using the slope and y-intercept.

Example

Graph f(x)=34x+6f\left(x\right)=-\dfrac{3}{4}x+6 using the slope and y-intercept.

Answer: The slope of this function is 34-\dfrac{3}{4} and the y-intercept is (0,6)(0,6) We can start graphing by plotting the y-intercept and counting down three units and right 44 units. The first stop would be (4,3)(4,3), and the next stop would be (8,0)(8,0). Screen Shot 2016-07-13 at 1.35.32 PM

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