1.1 — When Linear Breaks Down

The Problem: A Ball in the Air

You throw a ball straight up. A friend records the height at three moments:

Time (seconds)	Height (meters)
0	2
1	17
2	22

You already know how to handle data like this — find the rate of change and write a linear equation. So let's try it.

Between $t = 0$ and $t = 1$, the ball rises $15$ meters per second. That gives us the line:

$$y = 15t + 2$$

Check: at $t = 0$, $y = 2$. At $t = 1$, $y = 17$. Both correct.

But at $t = 2$, this line predicts $y = 32$. The actual height is $22$.

The line misses by $10$ meters. Something is wrong — not with the data, but with the assumption that a straight line is the right model.

The key question: If a line can't capture this pattern, what kind of expression can?

Why Lines Fail: The Rate of Change Isn't Constant

Let's look more carefully at what's happening between the data points.

Interval	Change in height	Rate of change
$t = 0$ to $t = 1$	$+15$	$15$ m/s
$t = 1$ to $t = 2$	$+5$	$5$ m/s

The ball is slowing down. The rate of change itself is changing.

Linear functions have one defining property: the rate of change is constant. The slope is the same everywhere. That's exactly what makes them simple — and exactly what makes them inadequate here.

When the rate of change isn't constant, a line can still fit two points, but it will bend away from reality everywhere else. We need something more flexible.

If this idea of "rate of change" feels fuzzy, here's a simple test: in a table of values with equally spaced inputs, a linear function produces the same difference between consecutive outputs every time. The moment those differences start varying, you're looking at something nonlinear.

Spotting Quadratic Patterns: The Second Differences Test

The rate of change is changing — but is it changing in a predictable way? Let's find out.

Add more data points and compute the differences between consecutive heights (these are called first differences):

$t$	Height	First difference
0	2
1	17	$+15$
2	22	$+5$
3	17	$-5$
4	2	$-15$

The first differences are $15, 5, -5, -15$. They aren't constant (not linear), but notice: each first difference is $10$ less than the one before.

Now compute the differences of the differences — the second differences:

First difference	Second difference
$15$
$5$	$-10$
$-5$	$-10$
$-15$	$-10$

The second differences are all $-10$. Constant.

This is the signature of a quadratic pattern:

For linear functions, the first differences are constant. For quadratic functions, the second differences are constant.

This is a simple, powerful test you can apply to any table of equally spaced data. If the first differences aren't constant but the second differences are, you're looking at a quadratic relationship.

Check your understanding

Before reading on, look at this table:

$x$	$y$
0	3
1	5
2	11
3	21
4	35

Is this data linear, quadratic, or neither? Compute the first and second differences to decide.

Check your answer

First differences: $2, 6, 10, 14$. Not constant, so not linear. Second differences: $4, 4, 4$. Constant! This is a **quadratic** pattern.

First Look at $y = ax^2 + bx + c$

We know this data follows a quadratic pattern. But what does a quadratic expression actually look like?

The most general form is:

$$y = ax^2 + bx + c$$

There are three coefficients — $a$, $b$, and $c$ — and this is not an accident. A quadratic has three degrees of freedom, which means we need exactly three pieces of information (say, three data points) to pin it down. A line has two coefficients ($y = mx + b$) and needs two points. A quadratic is one step more flexible.

Let's connect this back to our ball. We had three data points:

$(0, 2)$, $(1, 17)$, $(2, 22)$

Substituting each into $y = ax^2 + bx + c$:

• $t = 0$: $\quad a(0)^2 + b(0) + c = 2 \quad \Rightarrow \quad c = 2$
• $t = 1$: $\quad a(1)^2 + b(1) + 2 = 17 \quad \Rightarrow \quad a + b = 15$
• $t = 2$: $\quad a(4) + b(2) + 2 = 22 \quad \Rightarrow \quad 4a + 2b = 20$

From the last two equations: $4a + 2b = 20$ and $a + b = 15$. Subtracting twice the second from the first gives $2a = -10$, so $a = -5$ and $b = 20$.

The equation that fits our data is:

$$y = -5t^2 + 20t + 2$$

Let's verify with the point we didn't use yet — $t = 3$:

$$y = -5(9) + 20(3) + 2 = -45 + 60 + 2 = 17$$

This matches the table. The line gave us $y = 15(3) + 2 = 47$, which is absurdly wrong. The quadratic nails it.

What each coefficient tells us (a first look)

Don't try to memorize these yet — we'll spend the next section building intuition for them. For now, just notice:

• $c = 2$ is the height at $t = 0$ — the starting value. (Just like the $b$ in $y = mx + b$ is the $y$-intercept.)
• $a = -5$ is negative, which means the parabola opens downward. This makes physical sense: the ball goes up, then comes back down.
• $b = 20$ affects the initial rate of increase. We'll make this precise later.

Connection to what you know: A linear function $y = mx + b$ has two parameters and produces a straight line. A quadratic $y = ax^2 + bx + c$ has three parameters and produces a curve — specifically, a parabola. It's the simplest step up from a line. Every linear function is actually a special case of a quadratic where $a = 0$.

Reflection

Let's step back and name what just happened.

We started with a real situation — a ball in the air — and discovered that the tool we already had (linear equations) wasn't enough. The data told us this: the rate of change wasn't constant.

But we didn't just notice the failure — we found a pattern in how the rate of change was changing. Constant second differences pointed us to a new kind of expression: $y = ax^2 + bx + c$. Three data points, three unknowns, one equation that fits perfectly.

What we learned: Linear functions describe constant change. Quadratic functions describe change that itself changes at a constant rate. They're the natural next step when lines break down.

Practice

Problem 1 — Concrete. A table of values is given:

$x$	0	1	2	3
$y$	1	4	11	22

(a) Compute the first and second differences. Is this quadratic?

(b) Use the three points $(0, 1)$, $(1, 4)$, $(2, 11)$ to find $a$, $b$, and $c$.

(c) Verify your equation with the point $(3, 22)$.

Solution

(a) First differences: $3, 7, 11$. Second differences: $4, 4$. Constant — this is quadratic. (b) From $(0,1)$: $c = 1$. From $(1,4)$: $a + b + 1 = 4$, so $a + b = 3$. From $(2,11)$: $4a + 2b + 1 = 11$, so $4a + 2b = 10$. Subtracting twice the first: $2a = 4$, so $a = 2$, $b = 1$. The equation is $y = 2x^2 + x + 1$. (c) At $x = 3$: $2(9) + 3 + 1 = 22$. It checks out.

Problem 2 — Pattern. Without computing, decide: which of these tables is quadratic?

Table A:

$x$	0	1	2	3	4
$y$	0	1	4	9	16

Table B:

$x$	0	1	2	3	4
$y$	0	1	8	27	64

Solution

Table A: first differences $1, 3, 5, 7$; second differences $2, 2, 2$. Constant — **quadratic** ($y = x^2$). Table B: first differences $1, 7, 19, 37$; second differences $6, 12, 18$. Not constant — **not quadratic** (it's actually $y = x^3$).

Problem 3 — Debug. A student claims that the data below is quadratic because "the numbers are getting bigger faster."

$x$	0	1	2	3	4
$y$	1	2	4	8	16

Are they right? Why or why not?

Solution

First differences: $1, 2, 4, 8$. Second differences: $1, 2, 4$. The second differences are **not** constant, so this is not quadratic. "Getting bigger faster" is not enough — it must get bigger faster *at a constant rate*. This data is actually exponential ($y = 2^x$).

Problem 4 — Transfer. A small business tracks its total revenue over its first few months:

Month	1	2	3	4	5
Revenue ($)	500	1100	1900	2900	4100

(a) Is the revenue growing linearly, quadratically, or neither?

(b) If quadratic, find the equation using the first three data points.

Solution

(a) First differences: $600, 800, 1000, 1200$. Second differences: $200, 200, 200$. Constant — **quadratic growth**. (b) Using $(1, 500)$, $(2, 1100)$, $(3, 1900)$: - $a + b + c = 500$ - $4a + 2b + c = 1100$ - $9a + 3b + c = 1900$ Subtracting the first from the second: $3a + b = 600$. Subtracting the second from the third: $5a + b = 800$. So $2a = 200$, meaning $a = 100$, $b = 300$, $c = 100$. The equation is $y = 100t^2 + 300t + 100$. Check at $t = 4$: $1600 + 1200 + 100 = 2900$. Correct.

What's Next

We now have a new kind of expression — $y = ax^2 + bx + c$ — and we know it produces a curve, not a line. But what does this curve actually look like? What role does each coefficient play? In the next section, we'll study the simplest quadratic, $y = x^2$, and build a complete picture of its shape.