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The square function and the parabola

Level 6 · Ages 16–17

Quick recall· Let's wake up what you already know before we start.

Compute (-5)².

The whole lesson

Read it again any time, including while you work through the exercises.

Throw a ball toward a hoop and watch its path: it rises, peaks, then comes back down, tracing a rounded and perfectly symmetric curve. This curve is called a parabola, and it is the graphic signature of quadratics, that is, functions where x appears squared.

The basic model is the square function, f(x) = x². Compute a few values: f(-2) = 4, f(-1) = 1, f(0) = 0, f(1) = 1, f(2) = 4. Two things stand out. First, a number and its opposite have the same square, so the curve is symmetric about the vertical axis. Second, the values go down to 0 then climb back up: f is decreasing on ]-∞, 0], increasing on [0, +∞[, with a minimum of 0 reached at 0.

Every function of the form f(x) = ax² + bx + c, with a nonzero, keeps this parabola shape. Its lowest or highest point is called the vertex, and the vertical line through the vertex is the axis of symmetry: the parabola folds exactly onto itself there. The x-coordinate of the vertex comes from a formula worth knowing by heart: x = -b/(2a). To find the y-coordinate of the vertex, you then simply compute the image of that value.

The sign of a decides the orientation. If a > 0, the parabola opens upward, like a bowl: it goes down then up, and the vertex is a minimum. If a < 0, it opens downward, like a bell: it goes up then down, and the vertex is a maximum. Vertex, axis of symmetry, orientation: with these three keys, you can sketch any parabola in seconds.

The worked example, step by step

Study the square function f(x) = x²: symmetry, variations and vertex.

  1. f(-2) = 4, f(-1) = 1, f(0) = 0, f(1) = 1, f(2) = 4

    I compute a small table of values to see the shape of the curve.

  2. f(-x) = (-x)² = x² = f(x)

    A number and its opposite have the same image: the parabola is symmetric about the y-axis.

  3. Sur ]-∞ ; 0] : 4, 1, 0 : ça descend

    To the left of 0, as x increases, the squares get smaller: f is decreasing on ]-∞, 0].

  4. Sur [0 ; +∞[ : 0, 1, 4 : ça monte

    To the right of 0, the squares grow: f is increasing on [0, +∞[. The lowest point is reached at 0.

Result: The parabola of f(x) = x² has the origin (0, 0) as its vertex, which is a minimum, and the line x = 0 as its axis of symmetry.

Find the vertex and the variations of f(x) = x² - 4x + 1.

  1. a = 1, b = -4, c = 1

    I identify the coefficients of the form ax² + bx + c. Since a = 1 > 0, the parabola opens upward.

  2. x = -b/(2a) = -(-4)/(2 × 1) = 4/2 = 2

    I apply the vertex x-coordinate formula, watching the sign of b.

  3. f(2) = 2² - 4 × 2 + 1 = 4 - 8 + 1 = -3

    I get the y-coordinate of the vertex by computing the image of 2. The vertex is the point (2, -3).

  4. Décroissante sur ]-∞ ; 2], croissante sur [2 ; +∞[

    Upward parabola: f falls until the vertex then rises. The minimum of f is -3, reached at x = 2, and the axis of symmetry is the line x = 2.

Result: Vertex (2, -3): f is decreasing on ]-∞, 2], increasing on [2, +∞[, with a minimum of -3 at x = 2.