Advanced level

Quadratic function

A quadratic function f(x) = ax² + bx + c draws a parabola. Meet the vertex and the axis of symmetry, the three forms of the formula — general, canonical and factored — and how the sign of a decides between a minimum and a maximum.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

All formulas

  • General form

    f(x)=ax2+bx+c(a0)f(x) = ax^2 + bx + c \quad (a \neq 0)

    the graph is a parabola

  • Coordinates of the vertex

    p=b2a,q=Δ4a=f(p)p = -\frac{b}{2a}, \quad q = -\frac{\Delta}{4a} = f(p)

    (p, q) — the turning point of the parabola

  • Canonical form

    f(x)=a(xp)2+qf(x) = a(x - p)^2 + q

    the vertex is readable straight off it

  • Factored form

    f(x)=a(xx1)(xx2)f(x) = a(x - x_1)(x - x_2)

    exists only when Δ ≥ 0; shows the zeros

  • Axis of symmetry

    x=px = p

    the vertical line through the vertex

  • Range

    Rf=q,+) for a>0R_f = \langle q, +\infty) \ \text{for } a > 0

    for a < 0 it is (−∞, q⟩

A quadratic function is a function given by

f(x)=ax2+bx+c(a0)f(x) = ax^2 + bx + c \quad (a \neq 0)

Its graph is a parabola. The condition a0a \neq 0 is not a formality: were aa zero, the formula would collapse to bx+cbx + c, a linear function, and the graph would straighten into a line.

Arms, vertex, axis of symmetry

A parabola has one turning point — the vertex (p,q)(p, q) — and is symmetric about the vertical line through it:

−1012345−3−2−10123456xyx = 2(2, −1)x₁ = 1x₂ = 3f(x) = x² − 4x + 3
The parabola f(x) = x² − 4x + 3: vertex (2, −1), zeros 1 and 3, and the dashed axis of symmetry x = 2.

The coordinates of the vertex come from the formulas:

p=b2a,q=f(p)=Δ4ap = -\frac{b}{2a}, \qquad q = f(p) = -\frac{\Delta}{4a}

where Δ=b24ac\Delta = b^2 - 4ac — the same delta we computed for quadratic equations. The safest way to get the second coordinate is as f(p)f(p): the delta formula gives the same number but invites a sign slip.

The sign of aa decides which way the arms point:

−4−3−2−101234−5−4−3−2−1012345xya > 0a < 0
Arms up for a > 0 and down for a < 0. The larger the absolute value of a, the narrower the parabola.

Extremum and range

The vertex is the lowest or the highest point of the graph, so its second coordinate is the extreme value of the function:

  • a>0a > 0 — arms up, a minimum of value qq at the vertex, range q,+)\langle q, +\infty);
  • a<0a < 0 — arms down, a maximum of value qq at the vertex, range (,q(-\infty, q\rangle.

So a quadratic function never takes all real values — half of the vertical axis always stays empty. The domain, in contrast, is every real number: anything can be squared.

The axis of symmetry x=px = p is also the boundary of monotonicity. For a>0a > 0 the function decreases to the left of pp and increases to the right; for a<0a < 0 it is the other way round.

Find the vertex, the extremum and the range of f(x) = x² − 4x + 3.

The three forms

The same function can be written in three ways, and each shows something different without any computation:

ax2+bx+cgeneral=a(xp)2+qcanonical=a(xx1)(xx2)factored\underbrace{ax^2 + bx + c}_{\text{general}} = \underbrace{a(x - p)^2 + q}_{\text{canonical}} = \underbrace{a(x - x_1)(x - x_2)}_{\text{factored}}
  • the general one shows the value at zero: f(0)=cf(0) = c;
  • the canonical one shows the vertex: (p,q)(p, q);
  • the factored one shows the zeros: x1x_1 and x2x_2.

The factored form exists only when there are zeros at all, that is when Δ0\Delta \geq 0 (at Δ=0\Delta = 0 it takes the shape a(xx0)2a(x - x_0)^2). The canonical form exists always.

Write f(x) = x² − 4x + 3 in the canonical and the factored form.

Note that the vertex sits exactly halfway between the zeros: 1+32=2\frac{1 + 3}{2} = 2. That is the fastest route to pp once the roots are known — and it comes straight from the symmetry of the parabola.

The zeros

The zeros of a quadratic function are the solutions of ax2+bx+c=0ax^2 + bx + c = 0, that is the points where the parabola crosses the xx-axis. How many there are depends on the delta:

Δ>0two,Δ=0one,Δ<0none\Delta > 0 \Rightarrow \text{two}, \qquad \Delta = 0 \Rightarrow \text{one}, \qquad \Delta < 0 \Rightarrow \text{none}

The whole computation — the formulas for x1x_1 and x2x_2, the incomplete cases, Vieta's formulas — lives in the lesson on quadratic equations. What matters here is the consequence for the graph: when Δ<0\Delta < 0 the parabola lies entirely above the axis (for a>0a > 0) or entirely below it (for a<0a < 0), so the function has a constant sign.

Where you see it

A parabola is not a school ornament but a shape nature and engineering pick on their own:

  • projectile motion — a stone thrown at an angle travels along a parabola (air resistance aside);
  • satellite dishes and reflectors — a parabolic dish gathers parallel rays into a single point, the focus;
  • optimisation — questions like "the largest area for a given perimeter" lead straight to the vertex.
A 40 m fence is to enclose a rectangular plot. Which dimensions give the largest area?

Exercises

The session covers the four skills of this lesson: the value at a point, the coordinates of the vertex, and rewriting the general form into the canonical and the factored one. For the vertex type two numbers separated by a comma — first pp, then qq. For a change of form type the finished formula, e.g. (x-2)^2-1.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0
f(x) = x² + 5x + 6 → f(5)

Common mistakes

  • Losing the minus in pp — the formula is p=b2ap = -\frac{b}{2a}, so b=4b = -4 gives +2+2.
  • Confusing pp with qqpp lives on the horizontal axis, qq is a value of the function.
  • The sign of aa and the arms — the arms point up when a>0a > 0, not "always up".
  • A factored form at Δ<0\Delta < 0 — there is none, because there are no zeros.
  • Taking cc for a zerocc is the value at zero, the crossing with the yy-axis.
  • Calling the range all real numbers — a parabola always stops at qq.

Formula card

Topic: Quadratic function

  • General form

    f(x)=ax2+bx+c(a0)f(x) = ax^2 + bx + c \quad (a \neq 0)

    the graph is a parabola

  • Coordinates of the vertex

    p=b2a,q=Δ4a=f(p)p = -\frac{b}{2a}, \quad q = -\frac{\Delta}{4a} = f(p)

    (p, q) — the turning point of the parabola

  • Canonical form

    f(x)=a(xp)2+qf(x) = a(x - p)^2 + q

    the vertex is readable straight off it

  • Factored form

    f(x)=a(xx1)(xx2)f(x) = a(x - x_1)(x - x_2)

    exists only when Δ ≥ 0; shows the zeros

  • Axis of symmetry

    x=px = p

    the vertical line through the vertex

  • Range

    Rf=q,+) for a>0R_f = \langle q, +\infty) \ \text{for } a > 0

    for a < 0 it is (−∞, q⟩

−1012345−3−2−10123456xyx = 2(2, −1)x₁ = 1x₂ = 3f(x) = x² − 4x + 3
The parabola y = x² − 4x + 3 with vertex (2, −1), zeros 1 and 3, and the axis of symmetry x = 2.
−4−3−2−101234−5−4−3−2−1012345xya > 0a < 0
The sign of a decides which way the arms point: a > 0 — arms up and a minimum, a < 0 — arms down and a maximum.

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