IB Maths AI HL Further Functions & Graphs Paper 1 & 2 Vertex, axis, roots ~8 min read

Quadratic Functions & Graphs

Every quadratic y = ax² + bx + c draws a parabola. The sign of a picks the orientation (U or ∩); the axis of symmetry x = −b/(2a) locates the vertex; the discriminant tells you how many x-intercepts there are. With those four numbers you can sketch any parabola.

📘 What you need to know

Form: y = ax² + bx + c with a ≠ 0. a > 0 ⇒ U-shape, minimum; a < 0 ⇒ ∩-shape, maximum.
Axis of symmetry: x = −b/(2a) — in the formula booklet.
Vertex (turning point): lies on the axis of symmetry. x-coordinate is −b/(2a); y-coordinate by substitution.
y-intercept: (0, c) — read straight off the constant term.
Roots (x-intercepts, zeros): solutions of ax² + bx + c = 0. The discriminant b² − 4ac gives 2, 1, or 0 real roots.
If two real roots x₁, x₂ exist, the axis of symmetry is at their midpoint: x = (x₁ + x₂)/2.

Shape and orientation

The leading coefficient a determines the orientation. Positive a gives a U-shaped curve with a single minimum at the vertex; negative a gives an inverted ∩-shape with a single maximum. Either way the curve has a single vertical axis of symmetry running through the vertex.

Key features and how to find them

A complete sketch labels four things: the y-intercept (read off c), the axis of symmetry (use x = −b/(2a)), the vertex (substitute that x back in), and the x-intercepts if they exist. The discriminant b² − 4ac tells you in advance whether to look for two roots, one (a tangent at the x-axis), or none.

The axis of symmetry runs vertically through the vertex; the y-intercept is the constant term c; the roots (if any) sit symmetrically around the axis.

Quadratic graph (formula booklet) y = ax² + bx + c; axis of symmetry: x = −b2a discriminant b² − 4ac: > 0 (2 roots), = 0 (1 repeated root), < 0 (no real roots)

Sketching a quadratic, step by step

Once you’ve decided the orientation from the sign of a, the rest is mechanical: compute the axis of symmetry, drop that x into the equation for the vertex, set x = 0 for the y-intercept, and use the GDC or the quadratic formula for the roots. Plot those four labelled points and draw a smooth parabola through them.

Use the discriminant first if you only need to know how many roots there are — you don’t need to solve the equation to count them.

🧭 Recipe — sketching a quadratic

Read off the shape from the sign of a: positive ⇒ U, negative ⇒ ∩.
Axis of symmetry: x = −b/(2a).
Vertex: substitute that x-value back into the equation.
y-intercept: (0, c).
Roots: solve ax² + bx + c = 0 (factor, formula, or GDC). Plot all labelled points and draw the smooth parabola.

Worked examples

WE 1

Maximum or minimum?

Does the graph of y = −2x² + 5x + 3 have a maximum or a minimum?

look at the sign of a a = −2 < 0 a negative ⇒ ∩-shape maximum the orientation is settled before you do anything else.

WE 2

Vertex of a standard quadratic

Find the vertex of y = x² − 6x + 8.

axis of symmetry: x = −b/(2a) x = −(−6)/(2·1) = 3 substitute x = 3 to get y y = 9 − 18 + 8 = −1 vertex (3, −1) positive a means this is a minimum.

WE 3

Full feature list (two real roots)

For y = x² + 2x − 8, find the axis of symmetry, vertex, x-intercepts and y-intercept.

axis: x = −b/(2a) = −2/2 = −1 vertex: f(−1) = 1 − 2 − 8 = −9 vertex (−1, −9) roots: factor x² + 2x − 8 (x + 4)(x − 2) = 0 ⇒ x = −4, 2 y-intercept: (0, −8) axis x = −1 · vertex (−1, −9) · roots (−4, 0), (2, 0) · y-int (0, −8) trace check: roots average to (−4 + 2)/2 = −1 = axis ✓

WE 4

Quadratic with no real roots

For y = x² − 4x + 7, determine the number of roots and list the key sketch features.

discriminant first b² − 4ac = 16 − 28 = −12 < 0 no real roots ⇒ the curve doesn’t touch the x-axis axis: x = 4/2 = 2 vertex: f(2) = 4 − 8 + 7 = 3 vertex (2, 3) y-intercept (0, 7) opens upward · vertex (2, 3) · y-int (0, 7) · no x-intercepts vertex above the x-axis + opens up ⇒ whole curve sits above the x-axis.

WE 5

Inverted parabola

For y = −x² + 4x + 5, find the axis of symmetry, vertex, roots and y-intercept.

a = −1 ⇒ opens downward axis: x = −b/(2a) = −4/(−2) = 2 vertex: f(2) = −4 + 8 + 5 = 9 vertex (2, 9) — a maximum roots: −x² + 4x + 5 = 0 ⇒ x² − 4x − 5 = 0 (x − 5)(x + 1) = 0 ⇒ x = 5, −1 y-intercept: (0, 5) axis x = 2 · max (2, 9) · roots (−1, 0), (5, 0) · y-int (0, 5)

WE 6

Projectile motion (applied)

A ball is thrown vertically. Its height (m) after t seconds is h(t) = −5t² + 20t. (a) When is the ball at maximum height? (b) What is the maximum height? (c) When does it hit the ground?

(a) axis of symmetry: t = −b/(2a) t = −20/(2·(−5)) = −20/(−10) = 2 t = 2 seconds (b) h(2) h(2) = −5(4) + 20(2) = −20 + 40 = 20 max height = 20 m (c) h(t) = 0 −5t² + 20t = 0 ⇒ −5t(t − 4) = 0 t = 0 (launch) or t = 4 ground at t = 4 seconds

💡 Top tips

Sign of a first: U or ∩ settles the whole picture before any arithmetic.
Axis-of-symmetry formula x = −b/(2a) is in the booklet — never derive from scratch.
Discriminant before factoring: it tells you whether real roots exist, saving the effort if they don’t.
If two real roots are easy to read, the axis is the midpoint of the roots — a quick shortcut.
For applied problems, the vertex is the optimum (max profit, max height, min cost, etc.).

⚠ Common mistakes

Sign slip on b in x = −b/(2a) — if b is already negative, two minuses give a positive.
Reading the vertex x-coordinate as the whole vertex — substitute back for the y-value.
Saying “no roots” without checking the discriminant — sometimes the curve really does miss the x-axis.
Drawing the parabola through the y-intercept on the wrong side of the axis of symmetry — symmetry must be respected.
Forgetting orientation: a positive-a parabola has a minimum, not a maximum.

Next up: Cubic Functions & Graphs — one degree higher, up to three roots and two turning points.

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