Slope and Intercepts: Free Response

Match each equation against $y = mx + b$, where $m$ is whatever multiplies $x$ and $b$ is the constant.

In the first, the constant is $-7$, not $7$, so the crossing sits below the origin:

$y = 6x + (-7), \qquad m = 6, \qquad b = -7.$

The second lists its terms in the other order, so reorder it before reading anything:

$y = 9 - 4x = -4x + 9, \qquad m = -4, \qquad b = 9.$

The number written first is the constant here, and the slope is the number attached to $x$, minus sign included. Reading left to right instead would have taken the tilt from a number that is not touching $x$ at all.

The third has no constant term written, which means the constant is $0$:

$y = -\tfrac{3}{4}x + 0, \qquad m = -\tfrac{3}{4}, \qquad b = 0,$

so that line crosses at the origin. In all three cases the $y$-intercept is the point $(0, b)$, not the bare number $b$.

Take the two numbers the form needs in the order the picture hands them over.

First the constant, which costs no work at all: the line crosses the vertical axis three units below the origin, so $b = -3$.

Then the slope. Move along the line from that crossing to the next crossing of two gridlines it passes through, which is one unit right and two units up. A run of $1$ with a rise of $2$ gives

$m = \frac{2}{1} = 2.$

Put both numbers into $y = mx + b$:

$y = 2x - 3.$

Check it against a point the drawing agrees on. At $x = 2$ the equation gives $y = 2(2) - 3 = 1$, and the line does pass through the crossing two units right of the origin and one unit above it.

Two separate questions, and one line of the form settles each.

Why the constant is the crossing. The $y$-axis is the collection of points whose first coordinate is $0$, so the line meets it at whatever $y$ comes out when $x$ is $0$. Put that into the form:

$y = m(0) + b = b.$

The $mx$ term is wiped out whatever $m$ happens to be, which is why the crossing depends on $b$ alone, and why the crossing as a point is $(0, b)$.

Why the order does not matter. The expressions $9 - 4x$ and $-4x + 9$ are one sum written in its two possible orders, and addition gives the same result either way:

$9 + (-4x) = -4x + 9.$

So the two equations describe the same line, and the slope is whichever number is attached to $x$ in either of them. What the reordering changes is only which number your eye reaches first, and that is exactly why it is worth rewriting an equation into the form's order before reading it.

Slope-intercept form wants $y$ alone on one side with nothing multiplying it, so solve the equation for $y$.

Move the $x$-term across first, by subtracting $3x$ from both sides:

$4y = -3x + 24.$

The $y$ still carries a coefficient of $4$, so divide by $4$. Every term means both of them, not just the one that moved:

$y = -\tfrac{3}{4}x + 6.$

Now the reading is immediate: $m = -\tfrac{3}{4}$ and $b = 6$, so the $y$-intercept is the point $(0, 6)$.

A cheap check goes back to the original equation. At $x = 0$ it says $4y = 24$, so $y = 6$, which is the crossing just found.

The form hands over exactly what drawing a line needs: one point to start from, and a direction to leave it in.

The intercept is the free point, because it takes no calculation. Plot $(0, 6)$ on the vertical axis.

For the direction, read the slope as a rise over a run:

$m = -\tfrac{3}{4} = \frac{-3}{4}.$

A run of $4$ to the right comes with a rise of $-3$, which is three units down. Stepping from the crossing:

$(0, 6) \;\longrightarrow\; (0 + 4,\ 6 - 3) = (4, 3).$

Plot $(4, 3)$, lay a ruler across the two points and draw. The negative slope is what makes the line fall as it travels right. Stepping once more, to $(8, 0)$, is a cheap check that the first two points were placed correctly.

Slope-intercept form is a claim about the SHAPE of an equation, not about the numbers that appear in it, and an equation not yet in that shape is not making the claim.

Track the coefficient of $x$ through the work. It starts on the left of

$3x + 4y = 24,$

is carried across, which reverses its sign, and is then divided by $4$ along with everything else. Two separate steps acted on it, so the number that finally multiplies $x$ is neither the one written at the start nor its opposite.

The constant travels a shorter road and is changed all the same: the first step leaves it alone, and the second divides it by $4$. So the number on the right of the original equation is not the height of the crossing either.

The general point outlives the two numbers. Reading $m$ and $b$ is legitimate only once $y$ stands alone with a coefficient of $1$, because that is the only arrangement the form describes. Stopping one step early, at

$4y = -3x + 24,$

leaves an equation whose numbers are still halfway through the journey.

Each intercept comes from zeroing one coordinate, and which coordinate to zero is decided by which axis the line is crossing.

The $y$-axis is where $x = 0$, so put $x = 0$ into the equation:

$y = \tfrac{2}{3}(0) - 4 = -4,$

which gives the point $(0, -4)$. That is just the constant of the equation, exactly as slope-intercept form promises.

The $x$-axis is where $y = 0$, so set $y = 0$ and solve what is left:

$0 = \tfrac{2}{3}x - 4, \qquad \tfrac{2}{3}x = 4, \qquad x = 4 \cdot \tfrac{3}{2} = 6,$

which gives the point $(6, 0)$. Multiplying by the reciprocal is what clears the fraction; dividing by $\tfrac{2}{3}$ is the same move.

Neither answer is a bare number. An intercept is a place, so it is reported as a point, and each point carries its zero in the coordinate belonging to the axis being crossed.

A claim about EVERY line is destroyed by a single line that breaks it, so the job is to produce one, not to argue in general.

Take $y = 3x$. Its $y$-intercept comes from $x = 0$:

$y = 3(0) = 0,$

so that point is $(0, 0)$. Its $x$-intercept comes from $y = 0$:

$0 = 3x, \qquad x = 0,$

so that point is $(0, 0)$ too. One point, where the claim promised two, so the claim is false.

What makes it fail is the line passing through the origin. The origin is the one point lying on both axes, so a line through it crosses both axes in the very same place. This is a whole family of counterexamples rather than a freak case: $y = 3x$, $y = -x$ and $y = \tfrac{1}{5}x$ all do it.

The repaired claim is worth stating carefully, and it has to be checked in both directions. For a line with a nonzero slope, the two intercepts are different points exactly when the line does not pass through the origin. Forwards: if the line does pass through the origin then $b = 0$, and both rules land on $(0, 0)$, as above. Backwards: if the two intercepts are the same point, that point lies on both axes at once, and the origin is the only such point, so the line passes through the origin. Both directions hold, so this version is safe. The excluded cases are a slope of $0$, and a vertical line, which has no slope to speak of. Take the flat lines first: the line $y = 0$ meets the $x$-axis at every one of its points rather than at one, and $y = b$ with $b \neq 0$ never meets it at all, so neither has a single $x$-intercept to compare. The vertical lines are excluded by the same wording, since asking for a nonzero slope is already asking the line to have a slope, and $x = a$ has none.

Neither rule is about intercepts to begin with. Each one describes an axis, and the equation is only asked to supply the missing coordinate.

The $y$-axis is exactly the collection of points whose first coordinate is zero: $(0, -4)$, $(0, 7)$ and $(0, 0)$ all sit on it, and $(1, 0)$ does not. So to land on the $y$-axis you must set the first coordinate to zero, which is the demand $x = 0$, and the equation then tells you the only height that goes with it:

$y = m(0) + b = b.$

The $x$-axis is the mirror image, the points whose second coordinate is zero. To land on it you set $y = 0$, and the equation is left to be solved for $x$.

So the letter you set to zero is the letter that does NOT name the axis you are crossing, which sounds backwards until you see why: you zero the coordinate that the axis holds fixed. On the $y$-axis it is $x$ that is pinned at zero while $y$ ranges freely, which is precisely why $x = 0$ finds a crossing of that axis. Rebuilt that way, there is nothing left to remember.

Build the equation from the story one piece at a time.

The account starts with $45$ dollars, so at $t = 0$ the balance is $45$. Every day takes $5$ dollars away, so after $t$ days the amount charged is $5t$, and what is left is what there was minus what has been taken:

$D = 45 - 5t.$

That is the natural way to say it in words, and it is not yet in slope-intercept form, because the constant is written first. Reorder it:

$D = -5t + 45.$

Now match it against $y = mx + b$, with $t$ playing the part of $x$ and $D$ the part of $y$. The slope is the number multiplying $t$, and the intercept is the constant, so the graph crosses the vertical axis at $(0, 45)$.

Both numbers came straight out of the story: one is the balance before any day had been charged for, the other is what a single day costs the account, carrying the sign of the direction the balance moves.

The two rules are the same as for $x$ and $y$; only the letters have changed.

Set $t = 0$ to find where the graph meets the vertical axis:

$D = -5(0) + 45 = 45,$

which is the point $(0, 45)$. In the situation this is the balance at the instant of the top-up, so $45$ dollars.

Set $D = 0$ to find where it meets the horizontal axis:

$0 = -5t + 45, \qquad 5t = 45, \qquad t = 9,$

which is the point $(9, 0)$. Here $D = 0$ means no dollars left, so this is the day the credit runs out, $9$ days after the top-up.

The units are not decoration. The first coordinate of each point is a number of days and the second is a number of dollars, and without saying so, $(9, 0)$ is a pair of numbers rather than an answer to a question about an account.

The slope of a modelled line is a rate, and its sign is a direction.

Here the slope is $-5$, measured in dollars per day. Its size says the balance changes by $5$ dollars in each day, and its minus sign says the change is downward, which is what a charge does. A positive slope would have described an account being paid into.

Now the limit of the model. The line runs on forever in both directions; the account does not. One day past the crossing found in part B, the equation says

$D = -5(10) + 45 = -5,$

and a prepaid account carries no $-5$ dollars: once the credit reaches zero the plan simply stops. Nothing has gone wrong with the algebra. The equation was built from the sentence 'five dollars are taken every day', and that sentence stops being true once there is nothing left to take.

So the equation describes this account from the day of the top-up until the day the credit runs out, and the piece of the line worth drawing is the piece between the two intercepts. Beyond that the line is still a perfectly good line; it is just no longer a description of the account.

The letters change nothing about the method: clear the $x$-term, then remove the coefficient of $y$.

Subtract $Ax$ from both sides:

$By = -Ax + C.$

Now divide every term by $B$, which is precisely the step the hypothesis $B \neq 0$ is there to license:

$y = -\frac{A}{B}x + \frac{C}{B}.$

The equation is now in the shape $y = mx + b$, so the reading is immediate:

$m = -\frac{A}{B}, \qquad b = \frac{C}{B},$

and the $y$-intercept as a point is $\left(0, \tfrac{C}{B}\right)$.

Test it against a case already worked by hand. In $3x + 4y = 24$ the letters are $A = 3$, $B = 4$ and $C = 24$, and the formula returns a slope of $-\tfrac{3}{4}$ and an intercept of $\tfrac{24}{4} = 6$, which is what solving that equation line by line produced.

The $x$-intercept rule is to set $y = 0$, and it can be applied to the equation as it was handed over, with no conversion first.

Putting $y = 0$ into $Ax + By = C$ kills the $y$-term whatever $B$ may be:

$Ax + B(0) = C, \qquad Ax = C.$

Dividing by $A$ finishes it:

$x = \frac{C}{A},$

so the point is $\left(\tfrac{C}{A}, 0\right)$.

That last division is legitimate only when $A \neq 0$, and the condition is not a technicality. If $A = 0$ the equation reduces to $By = C$, that is, $y = \tfrac{C}{B}$, a line of slope $0$ sitting at one fixed height. Such a line meets the $x$-axis nowhere at all, unless that fixed height is itself zero, in which case the line IS the $x$-axis and meets it everywhere. Either way there is no single $x$-intercept to report, which is exactly what the impossible division was warning about.

Put $B = 0$ into the equation before arguing about anything, and see what is left standing:

$Ax + 0 \cdot y = C, \qquad Ax = C, \qquad x = \frac{C}{A}.$

The letter $y$ has vanished, so the equation places no restriction on it at all: any point whose first coordinate is $\tfrac{C}{A}$ satisfies it, whatever its second coordinate. That is a vertical line.

Now the claim to justify. Suppose such a line could be written as $y = mx + b$ for some numbers $m$ and $b$. The line contains both $\left(\tfrac{C}{A}, 0\right)$ and $\left(\tfrac{C}{A}, 1\right)$, and each would then have to satisfy that equation. The first gives

$0 = m \cdot \frac{C}{A} + b,$

and the second gives

$1 = m \cdot \frac{C}{A} + b.$

The right-hand sides are identical, so the left-hand sides must be equal too, which says $0 = 1$. That is false, so no such $m$ and $b$ exist, and the assumption $B \neq 0$ cannot be dropped: it is what keeps the vertical case out.

One piece of vocabulary is worth getting right at the close. A vertical line's slope is UNDEFINED, not zero. Zero is the slope of a horizontal line, the opposite picture, and the case $A = 0$ from part B. The two get swapped constantly, and the difference between them is exactly the difference between a line this form can describe and one it cannot.