Canonical form: the vertex of the quadratic
Learning Objectives
- Define the quadratic function \(f \colon \mathbb{R} \to \mathbb{R},\; f(x) = ax^2 + bx + c,\; a \neq 0\).
- Bring any quadratic function to canonical form \(f(x) = a\left(x + \tfrac{b}{2a}\right)^2 - \tfrac{\Delta}{4a}\).
- Read the vertex \(V(x_V, y_V)\) with \(x_V = -\tfrac{b}{2a}\), \(y_V = -\tfrac{\Delta}{4a}\) from the canonical form.
- Decide from the sign of \(a\) whether \(y_V\) is the minimum or the maximum value, and prove it.
- State and use the symmetry \(f(x_V - t) = f(x_V + t)\) and the axis of symmetry \(x = x_V\).
- Solve simple optimization problems by locating the vertex.
From equation to function
Unit 1 treated \(ax^2 + bx + c\) as something to set equal to zero and solve. Now we change the point of view, as we did with the linear function: the expression becomes a rule, assigning to every real number \(x\) the value \(f(x) = ax^2 + bx + c\). Questions about equations give way to questions about behaviour: how large can \(f(x)\) get? How small? Is there a pattern in its values?
Here is such a question. Unit 1's garden had perimeter \(20\) m and area \(21\) m². But among all rectangles with perimeter \(20\) m — which one encloses the largest area? With one side \(x\), the area is
\[A(x) = x(10 - x) = -x^2 + 10x,\]
a quadratic function. The equation-solving of Unit 1 cannot answer "what is the largest value?" — that is a question about the function as a whole. One rewrite will answer it, and the rewrite is an old friend: completing the square.
The quadratic function
A quadratic function (or second-degree function) is a function
\[f \colon \mathbb{R} \to \mathbb{R}, \qquad f(x) = ax^2 + bx + c,\]
where \(a, b, c \in \mathbb{R}\) and \(a \neq 0\).
As with the linear function, the condition \(a \neq 0\) keeps the degree honest: with \(a = 0\) the rule collapses to \(bx + c\), which is linear (or constant), not quadratic. Setting \(f(x) = 0\) recovers exactly the quadratic equation of Unit 1 — everything we proved there stands ready for use.
The canonical form
Every quadratic function \(f(x) = ax^2 + bx + c\) can be written, for all \(x \in \mathbb{R}\), as
\[f(x) = a\left(x + \frac{b}{2a}\right)^{\!2} - \frac{\Delta}{4a}, \qquad \Delta = b^2 - 4ac.\]
Writing \(x_V = -\dfrac{b}{2a}\) and \(y_V = -\dfrac{\Delta}{4a}\), this reads
\[f(x) = a\,(x - x_V)^2 + y_V.\]
This expression is called the canonical form of the quadratic function.
Factor \(a\) out of the two terms that contain \(x\), then complete the square inside the bracket, exactly as in Lesson 1 of Unit 1:
\[f(x) = a\left(x^2 + \frac{b}{a}x\right) + c = a\left(\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2}{4a^2}\right) + c,\]
using \(x^2 + \tfrac{b}{a}x = \left(x + \tfrac{b}{2a}\right)^2 - \tfrac{b^2}{4a^2}\). Distributing \(a\) and collecting the constants:
\[f(x) = a\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2}{4a} + c = a\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2 - 4ac}{4a},\]
and the last fraction is \(\dfrac{\Delta}{4a}\). Every step is an identity, valid for every real \(x\).
Vertex and axis of symmetry
For a quadratic function \(f(x) = ax^2 + bx + c\), the point
\[V\left(x_V,\; y_V\right) = \left(-\frac{b}{2a},\; -\frac{\Delta}{4a}\right)\]
is called the vertex of the function, and the vertical line \(x = x_V\) is called its axis of symmetry. Note that \(y_V = f(x_V)\): the vertex records an input together with its value.
One pleasant consequence falls out immediately: \(y_V = -\tfrac{\Delta}{4a}\) is zero exactly when \(\Delta = 0\) — the vertex sits at height \(0\) precisely when the equation \(f(x) = 0\) has its double root, and that root is \(x_V\) itself.
Extremum and symmetry
Let \(f(x) = a(x - x_V)^2 + y_V\) be a quadratic function in canonical form.
If \(a > 0\), then \(f(x) \geq y_V\) for all \(x \in \mathbb{R}\), with equality only at \(x = x_V\): the value \(y_V\) is the minimum of the function.
If \(a < 0\), then \(f(x) \leq y_V\) for all \(x \in \mathbb{R}\), with equality only at \(x = x_V\): the value \(y_V\) is the maximum of the function.
For every \(t \in \mathbb{R}\), \(f(x_V - t) = f(x_V + t)\): inputs placed symmetrically around \(x_V\) receive equal values, so the graph of \(f\) is symmetric with respect to the line \(x = x_V\).
All three claims flow from the same source: the square \((x - x_V)^2\) is \(\geq 0\) for every \(x\), and equals \(0\) only at \(x = x_V\).
If \(a > 0\), then \(a(x - x_V)^2 \geq 0\), so \(f(x) = a(x - x_V)^2 + y_V \geq y_V\), with equality exactly when the square vanishes — at \(x = x_V\). If \(a < 0\) the same term is \(\leq 0\) and the inequality reverses.
For the symmetry, evaluate at \(x_V + t\) and \(x_V - t\):
\[f(x_V + t) = a\,t^2 + y_V, \qquad f(x_V - t) = a\,(-t)^2 + y_V = a\,t^2 + y_V,\]
equal because \((-t)^2 = t^2\). Thus the points \((x_V - t,\; f(x_V - t))\) and \((x_V + t,\; f(x_V + t))\) of the graph are mirror images across the line \(x = x_V\), for every \(t\).
Reading a function from its canonical form
Take \(f(x) = x^2 - 4x + 3\). Completing the square:
\[f(x) = (x - 2)^2 - 1,\]
so \(x_V = 2\), \(y_V = -1\): the vertex is \(V(2, -1)\), the minimum value is \(-1\) (since \(a = 1 > 0\)), attained only at \(x = 2\), and the axis of symmetry is the line \(x = 2\).
Watch all of it happen in a strip of values:
\[f(0) = 3, \quad f(1) = 0, \quad f(2) = -1, \quad f(3) = 0, \quad f(4) = 3.\]
The values descend to \(-1\) at \(x = 2\) and then climb back — and they climb back through the same numbers: \(f(1) = f(3)\), \(f(0) = f(4)\). That is the symmetry \(f(2 - t) = f(2 + t)\) in action, one unit and two units away from the axis.
The best rectangle
Back to the opening question: among rectangles with perimeter \(20\) m, which area \(A(x) = x(10 - x) = -x^2 + 10x\) is largest? (Here \(0 < x < 10\), so that both sides are positive.)
With \(a = -1\), \(b = 10\), \(c = 0\):
\[x_V = -\frac{10}{2 \cdot (-1)} = 5, \qquad y_V = A(5) = 5 \cdot 5 = 25.\]
Since \(a < 0\), the vertex gives a maximum: the largest possible area is \(25\) m², attained only at \(x = 5\) — the rectangle with equal sides. Among all rectangles of a given perimeter, the square is the largest; no measuring or trial and error, just one completed square.
And Unit 1 fits neatly underneath: the area \(21\) of the original garden is below the maximum \(25\), which is why rectangles achieving it existed (sides \(3\) and \(7\)). Asking for area \(26\) would demand more than the maximum — no rectangle delivers it.
Connection to Computer Science
A guitar tuner, or an app that recognizes a song, measures the strength of a sound only at fixed frequency steps — yet it reports the pitch far more precisely than those steps. The standard trick: take the strongest measurement and its two neighbours, fit the unique quadratic function through those three points, and return its vertex \(x_V = -\tfrac{b}{2a}\) as the true location of the peak, in between the measured steps. This quadratic peak interpolation runs inside audio and radio software everywhere, thousands of times per second.
Two forms, one function
Is \(f(x) = x^2 - 4x + 3\) the same function as \(f(x) = (x - 2)^2 - 1\)? And which one is really \(f\)?
Recall the definition of a function from the general theory: a function is its domain, its codomain, and its assignment of values — not any particular formula. Two formulas define the same function exactly when they give equal values at every point of the domain. Here expanding \((x-2)^2 - 1 = x^2 - 4x + 3\) is an identity, so both formulas are the same function in different dress. Neither is more "really" \(f\) than the other.
Why keep several dresses in the wardrobe? Because each makes different facts visible at a glance. The general form shows \(f(0) = c\) and the coefficients that feed \(\Delta\) and Vieta. The canonical form shows the vertex, the extremum, and the symmetry. And Unit 1's factorization \(a(x - x_1)(x - x_2)\) — when \(\Delta \geq 0\) — shows the zeros. Three outfits, one function; skilled algebra is knowing which to wear for which occasion.
The habit of rigor here: to claim two formulas give the same function, verify the identity for all \(x\) — checking a few values is evidence, not proof.
Exercises
Exercise 1
For each function, compute the coordinates of the vertex \(V(x_V, y_V)\).
Exercise 2
For each function, choose its canonical form.
Exercise 3
Answer using the vertex — no value tables needed.
Exercise 4
Decide whether each statement about quadratic functions is true or false.
Exercise 5
From the sign of the leading coefficient, decide whether each function has a minimum or a maximum.
Summary
The quadratic function is \(f \colon \mathbb{R} \to \mathbb{R},\; f(x) = ax^2 + bx + c\) with \(a \neq 0\); its zeros are the solutions of Unit 1's equation.
Completing the square gives the canonical form \(f(x) = a(x - x_V)^2 + y_V\) with \(x_V = -\dfrac{b}{2a}\), \(y_V = -\dfrac{\Delta}{4a} = f(x_V)\); the point \(V(x_V, y_V)\) is the vertex.
Extremum: \(a > 0\) makes \(y_V\) the minimum value, \(a < 0\) the maximum — in both cases attained only at \(x = x_V\). This solves optimization problems outright (the best rectangle of given perimeter is the square).
Symmetry: \(f(x_V - t) = f(x_V + t)\) for all \(t\); the graph is symmetric about the axis of symmetry \(x = x_V\).
General, canonical, and factored form are three expressions of one function; each displays different information, and passing between them is pure identity algebra.