Calculate and Interpret Slope

In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, [latex]{x}_{1}[/latex] and [latex]{x}_{2}[/latex], and two corresponding values for the output, [latex]{y}_{1}[/latex] and [latex]{y}_{2}[/latex] which can be represented by a set of points, [latex]\left({x}_{1}\text{, }{y}_{1}\right)[/latex] and [latex]\left({x}_{2}\text{, }{y}_{2}\right)[/latex], we can calculate the slope [latex]m[/latex], as follows:

[latex]m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]

where [latex]\Delta y[/latex] is the change in output and [latex]\Delta x[/latex] is the change in input. In function notation, [latex]{y}_{1}=f\left({x}_{1}\right)[/latex] and [latex]{y}_{2}=f\left({x}_{2}\right)[/latex] so we could write:

[latex]m=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}[/latex]

The graph below indicates how the slope of the line between the points, [latex]\left({x}_{1,}{y}_{1}\right)[/latex] and [latex]\left({x}_{2,}{y}_{2}\right)[/latex], is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.

The slope of a function is calculated by the change in [latex]y[/latex] divided by the change in [latex]x[/latex]. It does not matter which coordinate is used as the [latex]\left({x}_{2,\text{ }}{y}_{2}\right)[/latex] and which is the [latex]\left({x}_{1},\text{ }{y}_{1}\right)[/latex], as long as each calculation is started with the elements from the same coordinate pair.

Point-Slope Form

Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Now we will learn another way to write a linear function called point-slope form which is given below:

[latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex]

where [latex]m[/latex] is the slope of the linear function and [latex]({x}_{1},{y}_{1})[/latex] is any point which satisfies the linear function.

The point-slope form is derived from the slope formula.

[latex]\begin{array}{llll}{m}=\frac{y-{y}_{1}}{x-{x}_{1}}\hfill & \text{assuming }{ x }\ne {x}_{1}\hfill\\{ m }\left(x-{x}_{1}\right)=\frac{y-{y}_{1}}{x-{x}_{1}}\left(x-{x}_{1}\right)\hfill & \text{Multiply both sides by }\left(x-{x}_{1}\right)\hfill\\{ m }\left(x-{x}_{1}\right)=y-{y}_{1}\hfill & \text{Simplify}\hfill \\ y-{y}_{1}={ m }\left(x-{x}_{1}\right)\hfill &\text{Rearrange}\hfill \end{array}[/latex]

Keep in mind that slope-intercept form and point-slope form can be used to describe the same linear function. We can move from one form to another using basic algebra. For example, suppose we are given the equation [latex]y - 4=-\frac{1}{2}\left(x - 6\right)[/latex] which is in point-slope form. We can convert it to slope-intercept form as shown below.

[latex]\begin{array}{llll}y - 4=-\frac{1}{2}\left(x - 6\right)\hfill & \hfill \\ y - 4=-\frac{1}{2}x+3\hfill & \text{Distribute the }-\frac{1}{2}.\hfill \\ \text{}y=-\frac{1}{2}x+7\hfill & \text{Add 4 to each side}.\hfill \end{array}[/latex]

Therefore, the same line can be described in slope-intercept form as [latex]y=-\frac{1}{2}x+7[/latex].

Point-slope form is particularly useful if we know one point and the slope of a line. For example, suppose we are told that a line has a slope of 2 and passes through the point [latex]\left(4,1\right)[/latex]. We know that [latex]m=2[/latex] and that [latex]{x}_{1}=4[/latex] and [latex]{y}_{1}=1[/latex]. We can substitute these values into point-slope form.

[latex]\begin{array}{l}y-{y}_{1}=m\left(x-{x}_{1}\right)\\ y - 1=2\left(x - 4\right)\end{array}[/latex]

If we wanted to rewrite the equation in slope-intercept form, we apply algebraic techniques.

[latex]\begin{array}{llll} y - 1=2\left(x - 4\right)\hfill & \hfill \\ y - 1=2x - 8\hfill & \text{Distribute the }2.\hfill \\ \text{}y=2x - 7\hfill & \text{Add 1 to each side}.\hfill \end{array}[/latex]

Both equations [latex]y - 1=2\left(x - 4\right)[/latex] and [latex]y=2x - 7[/latex] describe the same line. The graph of the line can be seen below.

Writing the Equation of a Line Using Two Points

Point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points [latex]\left(0,\text{ }1\right)[/latex] and [latex]\left(3,\text{ }2\right)[/latex]. We can use the coordinates of the two points to find the slope.

[latex]\begin{array}{l}{m}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \text{}{m}=\frac{2 - 1}{3 - 0}\hfill \\ \text{}{m}=\frac{1}{3}\hfill \end{array}[/latex]

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let’s use (0, 1) for our point.

[latex]\begin{array}{l}y-{y}_{1}=m\left(x-{x}_{1}\right)\\ y - 1=\frac{1}{3}\left(x - 0\right)\end{array}[/latex]

As before, we can use algebra to rewrite the equation in slope-intercept form.

[latex]\begin{array}{lll}y - 1=\frac{1}{3}\left(x - 0\right)\hfill & \hfill \\ y - 1=\frac{1}{3}x\hfill & \text{Distribute the }\frac{1}{3}.\hfill \\ \text{}y=\frac{1}{3}x+1\hfill & \text{Add 1 to each side}.\hfill \end{array}[/latex]

Both equations describe the line graphed below.

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