Given
Two points (days, rent) = (3, 285), (6, 510)
Find
A) An equation "in the standard form" for rent (y) in terms of days (x).
B) The equation of (A) in function notation.
C) The steps to graph the equation, mentioning lables and intervals.
Solution
A) A standard form equation can be written through point (h, k) using the idea that
... ∆y·x - ∆x·y = ∆y·h - ∆x·k . . . . . where ∆x and ∆y are the differences in x and y values, respectively
... ∆x = (6 - 3) = 3
... ∆y = 510 - 285 = 225
Then we can start with
... 225x -3y = 225·3 -3·285 = -180
To put this in "standard form", we must make the coefficients mutually prime. We can do that by dividing by 3.
... 75x -y = -60 . . . . . "standard form"
Some folks consider an equation in slope-intercept form to be "in the standard form". Here, that means solving for y.
... y = 75x + 60 . . . . . "in the standard form"
B) Function notation replaces y with f(x)
... f(x) = 75x + 60 . . . . . the equation in function notation
C) My favorite way to graph an equation of this sort is to type it into a graphing calculator, then adjust the axes so the portion of the graph of interest fills a reasonable portion of the screen. If the meaning of the variables is not made clear by the problem statement (or even if it is), their meaning and units can be entered as labels on the axes. Here, it is convenient to make the axes corresponding to the function domain be about [0, 8] or [0, 10] and the range about [0, 600]. These values cover the origin and the given points. (Usually, you want to show the origin on the graph, unless doing so compresses the data so much as to make the graph useless.)
I have labeled the axes as days (x) and dollars (y). The attached graph has grid lines at intervals of 50 vertically and 0.5 horizontally. That is what the graphing program sets them at. If you're using graph paper, you might want those to be 10 vertically and 0.2 horizontally for the smallest (minor) intervals.
To create the graph, plot the given points and draw a line through them.
