IB Maths AA HLTopic 2 — FunctionsPaper 1 & 2~9 min read
Quadratic Functions & Graphs
Straight lines were warm-up — now the curve appears. A quadratic function is just y = ax2 + bx + c, and its graph is a smooth U-shaped (or upside-down U-shaped) curve called a parabola. The whole game in this section is being able to look at a quadratic, picture its graph in your head, and pull out four key features: the shape, the y-intercept, the roots, and the vertex. Once you can do that, sketching becomes routine and most exam questions on quadratics are basically solved before you even start writing.
📘 What you need to know
A quadratic function has the general form f(x) = ax2 + bx + c, with a ≠ 0. Its graph is a parabola.
Shape: if a > 0 the parabola opens upward (∪) and has a minimum. If a < 0 it opens downward (∩) and has a maximum.
The y-intercept is at (0, c) — just read off the constant term.
The roots (or zeros, or x-intercepts) are the solutions to ax2 + bx + c = 0. A quadratic can have 0, 1 or 2 roots, decided by the discriminant (later note).
The axis of symmetry is the vertical line x = −b2a. This is in the formula booklet.
The vertex sits on the axis of symmetry. Its x-coordinate is −b2a; the y-coordinate is found by substituting into the function.
You’ll meet a quadratic written in three different forms:
• General: ax2 + bx + c (y-intercept obvious)
• Factorised: a(x − p)(x − q) (roots obvious)
• Vertex: a(x − h)2 + k (vertex obvious)
Pick the form that matches what the question gives you. They all describe the same parabola — they just highlight different features.
The shape — what the leading coefficient tells you
The first thing to look at in any quadratic is the sign of a, the coefficient of x2. That alone tells you whether the parabola opens up or down:
a > 0 — opens up (∪)
e.g. y = 2x2 − 3x + 1
graph has a minimum at the vertex; ends go up to +∞ on both sides
a < 0 — opens down (∩)
e.g. y = −x2 + 4x − 5
graph has a maximum at the vertex; ends drop to −∞ on both sides
The size of |a| controls how “narrow” or “wide” the parabola looks. A large |a| (like 5 or −10) gives a steep, narrow curve. A small |a| (like 0.2 or −13) gives a gentle, wide curve.
Before doing any algebra, glance at a and you’ve already pinned down the rough shape. Half the picture, for free.
The four key features of a parabola
Every parabola has the same four landmarks. Here’s what they look like on a graph:
Anatomy of a parabola — the four features that matter
The y-intercept — easiest one
The y-intercept is where the graph cuts the y-axis. That happens when x = 0. Plug x = 0 into ax2 + bx + c and you get y = c. So:
y-intercept
the y-intercept is at (0, c) — just read off the constant
For example, the curve y = 3x2 − 7x + 4 crosses the y-axis at (0, 4). No working needed.
The roots — where it cuts the x-axis
The roots (or zeros) are the x-values where y = 0 — i.e. where the curve crosses the x-axis. To find them, set the quadratic equal to zero and solve:
Roots
solve ax2 + bx + c = 0 for x
Three options for solving (covered in detail in upcoming notes):
Factorise
if it factors nicely
fastest for clean integer roots
Quadratic formula
x = −b ± √(b2 − 4ac)2a
always works (in formula booklet)
GDC
graph or solve menu
use on Paper 2 — quickest
A quadratic can have two distinct roots (cuts the x-axis twice), one repeated root (just touches the x-axis at the vertex), or no real roots (never reaches the x-axis). The discriminant b2 − 4ac tells you which case you’re in — there’s a whole separate note on that coming up.
The axis of symmetry and vertex — the two go together
A parabola is symmetric about a vertical line called the axis of symmetry. Both halves of the curve are mirror images of each other across this line. The formula:
Axis of symmetryx = −b2a✓ in the formula booklet
The vertex is the lowest (or highest) point of the parabola, and it lies right on the axis of symmetry. So:
🧭 Recipe — finding the vertex of ax2 + bx + c
Compute the x-coordinate: x = −b/(2a).
Substitute that x-value back into the function to get the y-coordinate.
Vertex is (x, y). Minimum if a > 0, maximum if a < 0.
Useful shortcut: if you already know the two roots p and q, the axis of symmetry is just halfway between them: x = (p + q)/2. Saves you computing −b/(2a) from scratch.
🤔 Why does x = −b/(2a) give the vertex?
Because the parabola is symmetric, the vertex sits exactly between the two roots. By the quadratic formula, the roots are x = (−b ± √D)/(2a) — symmetric about x = −b/(2a). The midpoint of the two roots is therefore exactly that value. The same line of symmetry holds even when the roots are complex, which is why the formula always works.
The three forms — pick the form that matches the question
A quadratic can be written three ways, all describing the same parabola. Each form puts a different feature in the spotlight:
General form
f(x) = ax2 + bx + c
y-intercept obvious: (0, c). Use when given three random points.
Factorised form
f(x) = a(x − p)(x − q)
roots obvious: (p, 0) and (q, 0). Axis of symmetry: x = (p+q)/2.
Vertex form
f(x) = a(x − h)2 + k
vertex obvious: (h, k). Axis of symmetry: x = h.
Vertex form has a bonus interpretation: it shows you how y = x2 has been transformed. Translate the basic parabola by (h, k), then vertically stretch by factor a — that gives you any quadratic. We’ll come back to this when we hit transformations.
Switching between the forms
The forms are interchangeable. Some practical conversions:
Factorised → general
expand the brackets
e.g. 2(x − 1)(x + 4) = 2x2 + 6x − 8
Vertex → general
expand the square
e.g. (x − 3)2 + 5 = x2 − 6x + 14
To go from general form to the other two, you’ll need either factorising (for factorised form) or completing the square (for vertex form) — both have their own notes coming up.
How to sketch a quadratic from scratch
An IB exam question often asks you to sketch a quadratic. Here’s the routine that works every time:
🧭 Recipe — sketching y = ax2 + bx + c
Shape: check the sign of a. Positive → ∪. Negative → ∩.
y-intercept: read off (0, c) and mark it on the y-axis.
Roots: solve ax2 + bx + c = 0. Mark them on the x-axis. (If no real roots, the curve doesn’t touch the x-axis.)
Vertex: x-coordinate is −b/(2a); substitute back to get the y-coordinate. Mark it.
Draw a smooth curve through all the labelled points, with the right shape, going through the vertex and continuing to the edges of your axes.
Always label the key points on your sketch — the y-intercept, the roots, and the vertex. A sketch without labels often gets you only half the marks. The graph from your GDC also counts as a labelled sketch on Paper 2 if you copy it accurately.
How to find the equation from given features
The reverse problem is also common: you’re given some info about a parabola and asked to find its equation. The trick is to start with the form that matches the info:
Given roots
use factorised
write y = a(x − p)(x − q); use a third point to find a
Given vertex
use vertex
write y = a(x − h)2 + k; use one more point to find a
Three random points
use general
write y = ax2 + bx + c; sub in to get a 3×3 system
Why factorised/vertex are better when applicable: they let you read off two of the three constants directly, leaving just one (a) to solve for. Going through general form means three unknowns and a system of three equations — much more work for the same answer.
Worked examples
WE 1
Read off key features from general form
For the quadratic function f(x) = 3x2 − 12x + 7, find: (a) the shape, (b) the y-intercept, (c) the axis of symmetry, (d) the coordinates of the vertex.
(a) Shape — look at sign of aa = 3 > 0 → opens upward (∪), has a minimum(b) y-intercept — read off cc = 7 → y-intercept at (0, 7)(c) Axis of symmetry — x = −b/(2a)x = −(−12) / (2 × 3) = 12 / 6 = 2(d) Vertex — substitute x = 2 into f(x)f(2) = 3(4) − 12(2) + 7 = 12 − 24 + 7 = −5∪-shape; (0, 7); x = 2; vertex (2, −5) (minimum)since a > 0 the vertex is the lowest point — the function never goes below y = −5
WE 2
Sketch a quadratic from general form
Sketch the graph of y = −x2 + 2x + 8, labelling the y-intercept, the roots, and the vertex.
Step 1: Shapea = −1 < 0 → ∩-shape (opens down)Step 2: y-interceptc = 8 → (0, 8)Step 3: Roots — solve −x² + 2x + 8 = 0multiply by −1: x² − 2x − 8 = 0factorise: (x − 4)(x + 2) = 0roots at x = −2 and x = 4 → (−2, 0) and (4, 0)Step 4: Vertexx = −b/(2a) = −2 / (2 × −1) = 1y = −(1)² + 2(1) + 8 = 9vertex (1, 9) — maximum since a < 0∩-shape with roots (−2, 0), (4, 0); y-intercept (0, 8); vertex/max (1, 9)check axis of symmetry: midpoint of roots = (−2 + 4)/2 = 1 ✓ matches x = 1
WE 3
Find the equation given the vertex and another point
A quadratic function has vertex (3, −4) and passes through the point (1, 0). Find its equation in the form y = a(x − h)2 + k.
Step 1: Vertex given → use vertex formy = a(x − 3)² + (−4)y = a(x − 3)² − 4Step 2: Substitute the other point (1, 0)0 = a(1 − 3)² − 40 = a(−2)² − 40 = 4a − 4a = 1y = (x − 3)² − 4check: at x = 1, y = (−2)² − 4 = 0 ✓; at the vertex x = 3, y = 0 − 4 = −4 ✓
WE 4
Find the equation given two roots and one more point
A quadratic graph cuts the x-axis at x = −2 and x = 5, and passes through the point (0, 20). Find its equation in the form y = ax2 + bx + c.
Step 1: Roots given → use factorised formy = a(x − (−2))(x − 5)y = a(x + 2)(x − 5)Step 2: Substitute (0, 20) to find a20 = a(0 + 2)(0 − 5)20 = a(2)(−5) = −10aa = −2Step 3: Expand to general formy = −2(x + 2)(x − 5)y = −2(x² − 3x − 10)y = −2x² + 6x + 20y = −2x² + 6x + 20check: y-intercept = 20 ✓; at x = −2, y = −8 − 12 + 20 = 0 ✓; at x = 5, y = −50 + 30 + 20 = 0 ✓
WE 5
Find the equation given three random points
A quadratic curve passes through the points (1, 6), (2, 5), and (4, 9). Find its equation in the form y = ax2 + bx + c.
Step 1: General form — substitute each point(1, 6): a + b + c = 6 …(i)(2, 5): 4a + 2b + c = 5 …(ii)(4, 9): 16a + 4b + c = 9 …(iii)Step 2: Eliminate c — (ii) − (i) and (iii) − (i)(ii) − (i): 3a + b = −1 …(iv)(iii) − (i): 15a + 3b = 3 → 5a + b = 1 …(v)Step 3: Solve for a, b(v) − (iv): 2a = 2 → a = 1sub into (iv): 3(1) + b = −1 → b = −4sub into (i): 1 + (−4) + c = 6 → c = 9y = x² − 4x + 9verify all three points: f(1) = 1 − 4 + 9 = 6 ✓, f(2) = 4 − 8 + 9 = 5 ✓, f(4) = 16 − 16 + 9 = 9 ✓
WE 6
Use the vertex to state the range of a quadratic
The quadratic function f(x) = 2x2 + 8x − 5 has domain x ∈ ℝ. Find the coordinates of the vertex and hence state the range of f.
Step 1: Find the x-coordinate of the vertexx = −b/(2a) = −8 / (2 × 2) = −2Step 2: Find the y-coordinate by substitutingf(−2) = 2(−2)² + 8(−2) − 5= 2(4) − 16 − 5= 8 − 16 − 5 = −13Step 3: Vertex (−2, −13). Since a = 2 > 0, this is a minimumfunction value never goes below −13vertex (−2, −13); range: f(x) ≥ −13, or [−13, ∞)the range of any ∪-quadratic is [k, ∞), where k is the y-coordinate of the minimum vertex; for a ∩-quadratic the range is (−∞, k]
💡 Top tips
Start by checking the sign of a. It tells you the shape (∪ or ∩) before you do any other work.
Memorise the four landmarks: shape, y-intercept (0, c), roots (solve = 0), vertex (−b/2a, then substitute). Every sketch question is just these four steps.
Pick the form that matches the question. Vertex given → vertex form. Roots given → factorised form. Random points → general form.
The axis of symmetry passes through the vertex and is exactly halfway between the two roots. Either formula gives the same answer.
Use your GDC on Paper 2. Graph mode shows roots and vertex instantly — no algebra needed. But always copy the values onto your sketch for marks.
For “find the range” questions, the vertex’s y-coordinate is the boundary. Range = [k, ∞) for ∪-shaped or (−∞, k] for ∩-shaped.
Always double-check by substituting points back in. If you found the equation from given features, the function should reproduce them exactly.
⚠ Common mistakes
Forgetting the negative sign in the axis-of-symmetry formula. It’s x = −b/(2a), not +b/(2a). Watch the sign carefully when b is already negative — you’ll get a positive answer.
Confusing the y-intercept with the vertex. They’re only the same point when the axis of symmetry happens to be the y-axis (i.e. b = 0). Otherwise they’re different points.
Forgetting a in factorised form. “Roots at p and q” doesn’t tell you the curve is (x − p)(x − q) — it could be any multiple of that. You always need a third point to find a.
Sign errors in vertex form. If the vertex is (−2, 5), the equation is a(x − (−2))2 + 5 = a(x + 2)2 + 5. The sign inside flips.
Solving “ax2 + bx + c” instead of “ax2 + bx + c = 0″ when looking for roots. The roots are where the function equals zero — set the expression equal to 0 first.
Missing the case of zero or one root. A parabola can totally avoid the x-axis (no real roots) or touch it just once (repeated root). Don’t assume there are always two roots to find.
Not labelling the sketch. Markers want to see the y-intercept, roots, and vertex labelled with their actual coordinates. An unlabelled curve is half-credit at best.
Quadratics are everywhere in IB — projectile motion in physics, profit functions in modelling, vertex of trajectories in calculus, the lot. Get really comfortable extracting the four features (shape, y-intercept, roots, vertex) from any quadratic in any of the three forms, and you’ll have a reliable foundation for everything that builds on this. The next note dives into factorising quadratics — the fastest method for finding roots when a quadratic factors cleanly. Then we’ll do completing the square, the quadratic formula, inequalities, and discriminants in turn.
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