Quadratic Functions & Parabolas

From equation to function

In the Algebra stratum, you solved quadratic equations — you found the $x$-values where $ax^2 + bx + c = 0$. Now we ask a bigger question: what does the entire function $f(x) = ax^2 + bx + c$ look like? What shape is the graph, and what does every coefficient do?

The answer is: the graph is a parabola, and the coefficient $a$ is its whole personality. One number. That's all it takes to tell me whether this thing is a cup or a cap, a smile or a frown.

$f(x) = ax^2 + bx + c, \qquad a \ne 0.$

• If $a > 0$: the parabola opens upward (cup shape, $\cup$). It has a minimum.
• If $a < 0$: the parabola opens downward (cap shape, $\cap$). It has a maximum.
• Larger $|a|$: narrower parabola. Smaller $|a|$: wider.

The parabola is symmetric — there's a vertical line through its lowest (or highest) point that is a mirror axis. That point is the vertex, and it's the single most important piece of information the parabola carries. Everything else is decoration.

Use the lab below to see how $a$, $b$, $c$ each transform the parabola:

Vertex form: the completing-the-square payoff

You completed the square to derive the quadratic formula — I know, I was there, the lab smelled like chalk and regret. Now watch the same move produce the vertex form.

Start from $f(x) = ax^2 + bx + c$. Factor $a$ from the first two terms:

$f(x) = a\!\left(x^2 + \frac{b}{a}x\right) + c.$

Complete the square inside the parentheses. The magic number to add (and subtract) is $\left(\frac{b}{2a}\right)^2$:

$f(x) = a\!\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c$ $= a\!\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c.$

Define $h = -\dfrac{b}{2a}$ and $k = c - \dfrac{b^2}{4a}$. Then:

$\boxed{f(x) = a(x - h)^2 + k.}$

This is vertex form. Why? Because $(x - h)^2 \ge 0$ always, and it equals $0$ exactly when $x = h$. So:

• The minimum (if $a > 0$) or maximum (if $a < 0$) output is $k$.
• It occurs at $x = h$.
• The vertex is $(h, k)$.

The vertex IS the function's personality in a single point. Everything else follows from it and $a$. If you remember nothing else from this lesson — though I expect you to remember everything — remember the vertex form and how to find it.

The axis of symmetry

The axis of symmetry is the vertical line $x = h = -\dfrac{b}{2a}$.

Why? The quadratic in vertex form, $a(x-h)^2 + k$, depends on $(x-h)^2$ — a squared quantity. Squaring is indifferent to sign: $(h + d)^2 = (h - d)^2$. So $f(h + d) = f(h - d)$ for any $d$. The function takes equal values at equal distances left and right of $h$.

Consequence: if the parabola has two real roots $r_1$ and $r_2$, they are symmetric around $x = h$. Their average is $h$:

$\frac{r_1 + r_2}{2} = h = -\frac{b}{2a}.$

(This is also Vieta's formula for the sum of roots: $r_1 + r_2 = -b/a$, so their average is $-b/2a$.)

The discriminant, drawn

You know the discriminant $\Delta = b^2 - 4ac$ from the Algebra stratum. Now see it geometrically:

• $\Delta > 0$: two distinct real roots. Parabola crosses the $x$-axis in two places.
• $\Delta = 0$: exactly one root (repeated). Parabola is tangent to the $x$-axis — the vertex sits exactly on it.
• $\Delta < 0$: no real roots. Parabola misses the $x$-axis entirely — vertex is strictly above (if $a > 0$) or below (if $a < 0$).

The discriminant tells you how many intersections the parabola has with the $x$-axis. That's it. Three answers; three pictures. Not six. Not twelve. Three. This should be the fastest check you do in the whole problem.

Max/min word problems: the first taste of optimization

The vertex gives the maximum or minimum output. Whenever a real-world problem asks "what is the greatest profit / shortest time / maximum area?" and the model is a quadratic, your job is exactly: find the damn vertex. Don't integrate. Don't differentiate. Just find the vertex — we built the machinery for exactly this.

Standard setup: fencing. You have $200$ meters of fencing to enclose a rectangular plot against a wall (so only three sides need fencing). If the side perpendicular to the wall has length $x$, the side parallel has length $200 - 2x$, and the area is:

$A(x) = x(200 - 2x) = -2x^2 + 200x.$

This is a downward parabola ($a = -2 < 0$), so it has a maximum. The vertex is at:

$x = -\frac{b}{2a} = -\frac{200}{2(-2)} = 50.$

Maximum area: $A(50) = 50 \cdot (200 - 100) = 50 \cdot 100 = 5000$ square meters.

This exact move — write the quantity as a quadratic, find the vertex — is the entire engine of optimization before calculus enters. File it. You'll need it more times than you expect.