Graphing Proportional Relationships

Proportional tables

Here is the walk as a table. Let $x$ be the number of hours walked and $y$ the number of miles covered, at a steady $3$ miles per hour.

Divide each $y$ by its $x$. You get $\frac{3}{1} = 3$, $\frac{6}{2} = 3$, $\frac{9}{3} = 3$ and $\frac{12}{4} = 3$: every ratio is the same number. In the language of the ratios chapter, the columns are equivalent ratios of miles to hours. The ratio $6 : 2$ is $3 : 1$ scaled by $2$, and $12 : 4$ is $3 : 1$ scaled by $4$.

That is what makes the table proportional. A quantity $y$ is proportional to a quantity $x$ when the ratio $\frac{y}{x}$ is the same nonzero number for every pair with $x$ not zero. Any pair with $x = 0$ must also have $y = 0$. That shared number is the constant of proportionality, written $k$. For the walk, $k = 3$. This lesson leaves out $k = 0$, the relationship where $y$ is always $0$.

The clause about $x = 0$ is there because that column has no ratio to test. The ratio $\frac{y}{0}$ has no value, whatever $y$ is, since division by zero has no answer. What the column must do instead follows from the ratio. If $y$ is always $k$ times $x$, then an input of $0$ gives $k \times 0 = 0$. Zero hours of walking covers zero miles.

So the test for a table has two parts. Divide $y$ by $x$ in every column where $x$ is not $0$, and the ratios must all match and not be $0$. Any column with $x = 0$ must have $y = 0$. If both hold, the table is proportional; if even one ratio differs, or $x = 0$ gives some other $y$, it is not.

The walk passes only because the pace is steady. A walker who stops for lunch covers no miles in that hour, and the ratios stop matching.

The equation y = kx

Start with one column of the walk. There $\frac{6}{2} = 3$, and multiplying both sides by $2$ gives $6 = 3 \times 2$: the miles are the rate times the hours. Every other column works the same way, since $9 = 3 \times 3$ and $12 = 3 \times 4$.

In general, every column with $x$ not zero has $\frac{y}{x} = k$. Multiply both sides by $x$, the balance move from solving one-step equations, which is safe because $x$ is not $0$:

The equation also covers the zero column, because $k \times 0 = 0$. So one equation describes the whole table. For the walk it is $y = 3x$.

The number $k$ is the unit rate. Put $x = 1$ into the equation and you get $y = k$, so $k$ is the output for one unit of input: the miles in one hour. That is the unit rate from the rates lesson, and it keeps its units. Write the walk’s constant as $k = 3$ miles per hour, not as a bare $3$. In general the units of $k$ are the units of $y$ per unit of $x$.

Which quantity you call $y$ decides which rate $k$ is. Put hours on top instead and the ratio is $\frac{1}{3}$ hour per mile. That is the same relationship read the other way round: the reciprocal rate, with the units reversed. So say which way you are reading it.

The equation answers questions in both directions. To find $y$, put in $x$ and multiply by $k$. To find $x$, put in $y$ and solve $kx = y$ by dividing both sides by $k$, which works because $k$ is not $0$.

Plotting a proportional table

Each column of a table is an ordered pair $(x, y)$. The walk gives $(0, 0)$, $(1, 3)$, $(2, 6)$, $(3, 9)$ and $(4, 12)$. Plot them with hours across and miles up.

The points line up exactly, and the line through them passes through the origin. Look at the step from each point to the next: it is $1$ across and $3$ up every time. That step is the unit rate drawn on the grid, one hour across and the $3$ miles walked in it up.

Once you have the line, you can read values that were never in the table. Halfway between $2$ and $3$ hours, at $x = 2.5$, the line sits at $7.5$ miles, and the equation agrees: $3 \times 2.5 = 7.5$. When a reading falls between gridlines, the equation gives the exact value and the picture gives a check.

Why the points line up through the origin

The walk’s points lined up, but that was one table. The claim is that this happens for every proportional relationship, whatever its $k$, and that the line always passes through the origin.

The reverse is also true, and it is what lets you test a graph by eye. Take the straight line through the origin and the point $(2, 6)$. The ratio at that point is $\frac{6}{2} = 3$. The proof above shows that the points of $y = 3x$ lie on one straight line. That line passes through the origin and through $(2, 6)$, since $3 \times 2 = 6$. Only one straight line passes through two given points, so the line we started with is the graph of $y = 3x$.

The same argument works for any straight line through the origin, other than an axis. Pick a point on it other than the origin, and call the ratio $\frac{y}{x}$ there $k$. The graph of $y = kx$ is a straight line through the origin and that point. Only one straight line passes through both, so the line is $y = kx$. The vertical axis is the exception, because every point on it has $x = 0$ and no ratio $\frac{y}{x}$ exists. The horizontal axis is left out too, because its $k$ would be $0$.

Steepness and the sign of k

Every line $y = kx$ passes through the origin and through $(1, k)$. Between two positive values of $k$, the larger one puts that second point higher. Its line climbs more for each step across, so it is steeper. A walk at $3$ miles per hour leaves a steeper trail than a walk at $2$. For a positive $k$, steepness on these graphs is the unit rate.

The rate can also be negative. Then $y$ goes down as $x$ goes up. The line $y = -2x$ passes through $(1, -2)$. From the origin, each step of $1$ across goes $2$ down, so the line falls from left to right. The ratio $\frac{y}{x}$ is still the same at every point except the origin: $\frac{-2}{1}$, $\frac{-4}{2}$ and $\frac{6}{-3}$ all equal $-2$.

So the sign of $k$ gives the direction, rising for a positive $k$ and falling for a negative one. The size of $k$, its absolute value, gives the steepness. The lines $y = 2x$ and $y = -2x$ are equally steep, tilted opposite ways.

Proportional or not?

A relationship can fail to be proportional in two different ways, and each shows up on the graph.

Take a bike rental that charges a $4$-dollar fee plus $2$ dollars per hour. One hour costs $6$ dollars, two cost $8$, three cost $10$ and four cost $12$. The cost goes up by the same $2$ dollars every hour, but the ratios are $6$, $4$, $\frac{10}{3}$ and $3$, which do not match. On the graph the points lie on a straight line, but the line meets the y-axis at $(0, 4)$, the fee, instead of at the origin.

Now take the area of a square: side $1$ gives area $1$, side $2$ gives $4$, side $3$ gives $9$. Here the ratios are $1$, $2$ and $3$. The graph does start at the origin, since a square with side $0$ has no area. But the points bend upward instead of lying on a straight line.

So the graph test has two parts. The points must lie on a straight line, and that line must pass through the origin (and not be either axis). Neither part is enough alone: the rental passes the first and fails the second, and the square passes the second and fails the first. In a table, the same two failures show up as ratios that do not match.

What the situation allows

The equation $y = 2x$ makes sense for every $x$: negatives, fractions and decimals included. Its full graph is a straight line through the origin, running in both directions. A real situation usually allows only some of those inputs, and its graph keeps only that part of the line.

Pencils at a fixed $2$ dollars each, with no bulk discount, follow $y = 2x$. You can buy $0$, $1$, $2$ or $3$ pencils, but not $2.5$ of one or $-1$. So the graph of the pencil situation is a row of separate dots at whole-number values of $x$. Every one of them sits on the line $y = 2x$.

A walk at a steady $2$ miles per hour also follows $y = 2x$. But a walk can last any length of time from $0$ up, $1.5$ hours included. Its graph is a solid piece of the line, starting at the origin and running to the right. Negative times make no sense, so the left half of the line is not part of it. A situation can also stop: a $3$-hour walk ends at the point $(3, 6)$.

Common mistakes