Quadratic Functions and Parabolas

What makes a function quadratic

A quadratic function is a function of the form

where $a$, $b$, and $c$ are real constants: $a$ is the leading coefficient, $b$ the linear coefficient, and $c$ the constant term. Squaring, multiplying by a constant, and adding accept every real input, so the domain is all real numbers. The graph of a quadratic function is called a parabola.

The condition $a \ne 0$ is not decoration. Set $a = 0$ and the rule collapses to $f(x) = bx + c$, a linear function: a straight line, with no bend, no highest or lowest point, and no vertical mirror. Everything this chapter studies lives in the $x^2$ term. The other two coefficients are optional in a way $a$ is not: $f(x) = x^2$ has $b = c = 0$ and is still the most important quadratic of all.

There is a precise sense in which quadratics are the simplest functions after linear ones. A linear function changes by equal amounts over equal steps: each step of $1$ in the input adds the same $b$ to the output. For $f(x) = x^2$ the outputs at $x = 0, 1, 2, 3, 4$ are $0, 1, 4, 9, 16$, and the jumps between them are $1, 3, 5, 7$: not constant, but growing by the same $2$ each time. That is no accident of these five values, because

so the jump is itself a linear function of $x$. The same computation on a general quadratic gives $f(x + 1) - f(x) = 2ax + a + b$, again linear. A linear function has constant jumps; a quadratic has linear jumps. Quadratics are what you get when you let the rate of change itself change, but only in the plainest possible way. The steady growth of those jumps is exactly the bend you see in a parabola.

One more piece of vocabulary before we begin. This lesson is about the quadratic function, the input-output object with a graph. Asking which inputs produce one prescribed output, that is, solving a quadratic equation, begins in the next lesson. Today we learn everything a parabola will tell us without solving anything.

The parent of every parabola

Transformations of Graphs had one message: complicated graphs are simple graphs moved around, and everything about the copy is readable from the original plus the move. For parabolas the original is $p(x) = x^2$, the parent parabola, so before transforming it you should know it completely.

Vertex form is a transformation in disguise

Now dress the parent. Take constants $a \ne 0$, $h$, and $k$, and build

Read the right side with chapter 2 eyes: the input is shifted by $h$ before the parent acts, and the output is scaled by $a$ and shifted by $k$ after. This is the master shape $a\,f(b(x - h)) + k$ from Transformations of Graphs with $b = 1$, so its landing rule applies verbatim: the point $(u, v)$ on the parent lands at $(u + h,\ av + k)$.

Everything you might want to know about $g$ now costs nothing, because you know where the parent’s features land. The parent’s vertex $(0, 0)$ lands at $(h, k)$, so the vertex of $g$ is $(h, k)$. The parent’s mirror $x = 0$ is carried to the vertical line through the new vertex, so the axis of symmetry is $x = h$. The parent’s range $[0, \infty)$ is scaled by $a$ and shifted by $k$. When $a > 0$ the range becomes $[k, \infty)$, and when $a < 0$ the negative scaling flips the interval end for end, giving $(-\infty, k]$. So $g$ has a minimum value of $k$ when $a > 0$ and a maximum value of $k$ when $a < 0$, attained at $x = h$ in both cases. Meanwhile, $|a|$ sets how steeply the arms climb: the parabola looks narrower for $|a| > 1$ and wider for $0 < |a| < 1$.

If you prefer not to lean on chapter 2, the same facts fall out of two lines of algebra. Since $(x - h)^2 \ge 0$ with equality exactly at $x = h$, multiplying by $a > 0$ and adding $k$ gives $g(x) \ge k$ with equality exactly at $x = h$; for $a < 0$ the inequality reverses. And equal steps either side of $h$ give equal heights, because $g(h + t) = at^2 + k = g(h - t)$, so the line $x = h$ is a genuine mirror. The transformation reading is instant, the direct check is airtight, and you should own both.

Both of those are states of the figure below, so check yourself against it rather than against the answer key. Worked example 1 is $a = -2$, $h = -3$, $k = 8$; the checkpoint is $a = -3$, $h = -2$, $k = 7$. Set each in turn and confirm that the vertex arrives where you said, and that the parabola opens the way the sign of $a$ promised.

The claim worth testing here is the one this section makes about the range, because it is the claim students take on trust. Fix $a$ at some positive value and move $k$: the lowest point of the curve is $k$, every time, and no part of the graph ever appears below it. Now make $a$ negative with $k$ left where it is. The same number is suddenly the highest point, and the curve hangs beneath it. Nothing about $k$ changed; the interval turned over because the scaling did. Moving $h$ instead slides the whole picture sideways and leaves that number exactly where it was, which is the other half of the claim. The range is fixed by $a$ and $k$ alone, and the axis of symmetry rides along with $h$.

What standard form shows at a glance

Multiply a vertex form out and you always land back in standard form: expanding $a(x - h)^2 + k$ gives $ax^2 - 2ahx + (ah^2 + k)$, an $ax^2 + bx + c$. Standard form is the costume quadratics usually arrive in, so it is worth knowing exactly what it offers for free, and what it hides.

Two features are free. First, the $y$-intercept: $f(0) = a \cdot 0 + b \cdot 0 + c = c$, so the graph crosses the $y$-axis at height $c$, no work required. Second, the end behavior. For $x \ne 0$, factor out the highest power, exactly as the Graphs of Functions lesson taught:

Far from the origin the fractions $\tfrac{b}{x}$ and $\tfrac{c}{x^2}$ shrink toward $0$, so the bracket sits as close to $a$ as you like while $x^2$ is enormous and positive. The outputs therefore take the sign of $a$ and grow without bound: when $a > 0$ both ends of the parabola point up, and when $a < 0$ both ends point down. One sign settles the whole large-scale picture.

What standard form hides is everything about the middle: where the turning point sits, how low or high the graph reaches, whether it ever meets the $x$-axis. The vertex is simply not visible in $ax^2 + bx + c$. The surprise of the next section is that symmetry digs it out anyway.

The axis of symmetry, from symmetry alone

Every parabola you have ever drawn has an evident vertical mirror, and for vertex form we just proved it: the mirror is $x = h$. But a quadratic handed to you as $f(x) = ax^2 + bx + c$ names no $h$. Later in this chapter, Completing the Square converts standard form into vertex form outright. Remarkably, you do not have to wait for it: two cheap evaluations and one symmetry principle locate the mirror exactly.

The verification handed us more than we asked for. The identity $f(m + t) = f(m) + at^2$ says the height at displacement $t$ from the axis differs from the height on the axis by $at^2$, a correction with one sign. That correction is positive for every $t \ne 0$ when $a > 0$, negative when $a < 0$, and it grows steadily as $|t|$ grows. So the point of the parabola on the axis is the lowest point of the whole graph when $a > 0$ and the highest when $a < 0$. In other words, the vertex of $f(x) = ax^2 + bx + c$ sits at

and the range is $\left[f\!\left(-\tfrac{b}{2a}\right), \infty\right)$ when $a > 0$ and $\left(-\infty, f\!\left(-\tfrac{b}{2a}\right)\right]$ when $a < 0$. Standard form never displays its vertex, but it cannot hide it either: one division finds the axis, one evaluation finds the height. When Completing the Square arrives, it will repackage exactly this information as a rewriting of the formula. The geometry was never waiting on the algebra.

One quadratic, three normal forms

You have now met all three costumes this chapter lives in:

The same $a$ appears in all three, because expanding either alternative form starts $ax^2 + \cdots$, so the leading coefficient, and with it the end behavior, survives every change of costume.

Each form pays out its own facts at sight. Standard form shows the $y$-intercept $c = f(0)$ and the end behavior. Vertex form shows the vertex $(h, k)$, the axis $x = h$, the extreme value $k$, the range, and the whole graph as a transformation of the parent. Factored form shows the zeros: substituting $x = r_1$ or $x = r_2$ makes a factor vanish, so $f(r_1) = f(r_2) = 0$ and the graph meets the $x$-axis at exactly those inputs. Reading zeros off a factored form that is handed to you is mere evaluation; producing the factored form yourself is the next lesson’s work. Factored form also controls the sign of $f$ between and beyond the zeros, which is what Quadratic Inequalities will exploit.

The three forms are not equally available. Every quadratic has a standard form, and Completing the Square will prove every quadratic has a vertex form. But the factored form exists over the real numbers only when the parabola actually meets the $x$-axis. A parabola that opens upward with its vertex above the axis, such as one with range $[4, \infty)$, never reaches height $0$. Such a parabola has no real zeros, and that is a complete, honest answer. The Quadratic Formula lesson builds the discriminant, the single number that decides in advance which situation you are in. And chapter 5 enlarges the number system so that even those parabolas acquire zeros.

Notice also what symmetry says about factored form: the two zeros are a pair of inputs sharing the output $0$, so the axis of symmetry must cut their midpoint, $x = \tfrac{r_1 + r_2}{2}$. Read backward, the coefficients remember the zeros, which is the seed of the Sum and Product of Roots lesson closing this chapter.

The chapter’s plan is now one sentence per lesson. Solving Quadratics by Factoring converts standard form to factored form. Completing the Square converts standard form to vertex form. The Quadratic Formula runs that second conversion once, in full generality, and keeps the result forever, which is why it is a theorem rather than a recipe. Quadratic Inequalities harvests the sign information the factored form displays. Sum and Product of Roots reads the factored form back into the coefficients, extracting facts about the zeros without ever finding them.

Common mistakes