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1.2 Basic classes of functions  (Page 2/28)

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( a + b ) − b 1 − 0 = a .

We have shown that the coefficient a is the slope of the line. We can conclude that the formula f ( x ) = a x + b describes a line with slope a . Furthermore, because this line intersects the y -axis at the point ( 0 , b ) , we see that the y -intercept for this linear function is ( 0 , b ) . We conclude that the formula f ( x ) = a x + b tells us the slope, a , and the y -intercept, ( 0 , b ) , for this line. Since we often use the symbol m to denote the slope of a line, we can write

f ( x ) = m x + b

to denote the slope-intercept form    of a linear function.

Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function passes through the point ( x 1 , y 1 ) and the slope of the line is m . Since any other point ( x , f ( x ) ) on the graph of f must satisfy the equation

m = f ( x ) − y 1 x − x 1 ,

this linear function can be expressed by writing

f ( x ) − y 1 = m ( x − x 1 ) .

We call this equation the point-slope equation    for that linear function.

Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the slope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be expressed in either of these forms. Instead, a vertical line is described by the equation x = k for some constant k . Since neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation

a x + b y = c ,

where a , b are both not zero, to denote the standard form of a line .

Definition

Consider a line passing through the point ( x 1 , y 1 ) with slope m . The equation

y − y 1 = m ( x − x 1 )

is the point-slope equation for that line.

Consider a line with slope m and y -intercept ( 0 , b ) . The equation

y = m x + b

is an equation for that line in slope-intercept form .

The standard form of a line is given by the equation

a x + b y = c ,

where a and b are both not zero. This form is more general because it allows for a vertical line, x = k .

Finding the slope and equations of lines

Consider the line passing through the points ( 11 , −4 ) and ( −4 , 5 ) , as shown in [link] .

An image of a graph. The x axis runs from -5 to 12 and the y axis runs from -5 to 6. The graph is of the function that is a decreasing straight line. The function has two points plotted, at (-4, 5) and (11, 4).
Finding the equation of a linear function with a graph that is a line between two given points.
  1. Find the slope of the line.
  2. Find an equation for this linear function in point-slope form.
  3. Find an equation for this linear function in slope-intercept form.
  1. The slope of the line is
    m = y 2 − y 1 x 2 − x 1 = 5 − ( −4 ) −4 − 11 = − 9 15 = − 3 5 .
  2. To find an equation for the linear function in point-slope form, use the slope m = −3 / 5 and choose any point on the line. If we choose the point ( 11 , −4 ) , we get the equation
    f ( x ) + 4 = − 3 5 ( x − 11 ) .
  3. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for f ( x ) . When we do this, we get the equation
    f ( x ) = − 3 5 x + 13 5 .
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Consider the line passing through points ( −3 , 2 ) and ( 1 , 4 ) . Find the slope of the line.

Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.

m = 1 / 2 . The point-slope form is

y − 4 = 1 2 ( x − 1 ) .

The slope-intercept form is

y = 1 2 x + 7 2 .

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A linear distance function

Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.

  1. Describe the distance D (in miles) Jessica runs as a linear function of her run time t (in minutes).
  2. Sketch a graph of D .
  3. Interpret the meaning of the slope.
  1. At time t = 0 , Jessica is at her house, so D ( 0 ) = 0 . At time t = 78 minutes, Jessica has finished running 9 mi, so D ( 78 ) = 9 . The slope of the linear function is
    m = 9 − 0 78 − 0 = 3 26 .

    The y -intercept is ( 0 , 0 ) , so the equation for this linear function is
    D ( t ) = 3 26 t .
  2. To graph D , use the fact that the graph passes through the origin and has slope m = 3 / 26 .
    An image of a graph. The y axis is labeled “y, distance in miles”. The x axis is labeled “t, time in minutes”. The graph is of the function “D(t) = 3t/26”, which is an increasing straight line that starts at the origin. The function ends at the plotted point (78, 9).
  3. The slope m = 3 / 26 ≈ 0.115 describes the distance (in miles) Jessica runs per minute, or her average velocity.
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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