The same quadratic can be written three ways, each highlighting a different feature. The GDC works with any form, but spotting the right form saves time.
Find a quadratic from features: given roots
p,
q and one other point, plug into
y =
a(
x −
p)(
x −
q) and solve for
a. Given vertex (
h,
k) and one other point, use vertex form instead.
🧭 Recipe — sketching any quadratic
- Read off the shape: if a > 0, the curve is ∪ (min); if a < 0, it’s ∩ (max).
- y-intercept: substitute x = 0 (or just read off the constant c). Plot (0, c).
- Find the roots: solve ax² + bx + c = 0 using the GDC (or factorising if it’s clean). Plot each root on the x-axis.
- Axis of symmetry & vertex: xvertex = −b/(2a). Substitute back into the equation to get yvertex. Plot the vertex (and the dashed axis if asked).
- Join up with a smooth curve: through the labelled points, in the correct shape (∪ or ∩). Label every key point with coordinates.
Worked examples
WE 1Standard form — find every key feature
For the quadratic y = 2x² + 8x − 10, find the axis of symmetry, the vertex, the y-intercept and the roots.
Step 1 — identify a, b, c
a = 2, b = 8, c = −10
Step 2 — axis of symmetry (formula booklet)
x = −b/(2a) = −8/4 = −2
Step 3 — vertex: substitute x = −2
y = 2(4) + 8(−2) − 10 = 8 − 16 − 10 = −18
vertex (−2, −18)
Step 4 — y-intercept (set x = 0)
y = −10 → (0, −10)
Step 5 — roots (factor or GDC)
2x² + 8x − 10 = 2(x² + 4x − 5)
= 2(x + 5)(x − 1) = 0
x = −5 or x = 1
axis x = −2 · vertex (−2, −18) · y-int (0, −10) · roots −5, 1
cross-check: midpoint of roots = (−5 + 1)/2 = −2 = axis ✓. Always confirm using this shortcut when you have both roots.
WE 2Negative leading coefficient — maximum
The quadratic y = −x² + 6x − 5 has a maximum point. Find the coordinates of that maximum and the roots.
Step 1 — a = −1 (negative ⇒ opens downward, has max)
Step 2 — axis of symmetry
x = −6/(2 × −1) = −6/−2 = 3
Step 3 — maximum: substitute x = 3
y = −(9) + 6(3) − 5 = −9 + 18 − 5 = 4
maximum (3, 4)
Step 4 — roots: solve −x² + 6x − 5 = 0
multiply by −1: x² − 6x + 5 = 0
(x − 1)(x − 5) = 0
x = 1 or x = 5
maximum (3, 4) · roots x = 1, x = 5
notice the maximum y-value (4) is positive even though the curve has roots — expected, since the curve has to climb above the x-axis to make a peak.
WE 3Factored form — read off the roots
The quadratic y = (x − 2)(x + 6) is given. State the roots, then find the axis of symmetry, the vertex, the y-intercept and the equation in standard form.
Step 1 — roots read straight from factored form
x − 2 = 0 or x + 6 = 0
x = 2 or x = −6
Step 2 — axis of symmetry = midpoint of roots
x = (2 + (−6))/2 = −4/2 = −2
Step 3 — vertex: substitute x = −2
y = (−2 − 2)(−2 + 6) = (−4)(4) = −16
vertex (−2, −16)
Step 4 — y-intercept and standard form
at x = 0: y = (−2)(6) = −12 → (0, −12)
expand: (x − 2)(x + 6) = x² + 4x − 12
roots −6, 2 · axis x = −2 · vertex (−2, −16) · y-int (0, −12)
factored form gives roots for free — no GDC needed. The midpoint shortcut for the axis is even faster than the −b/(2a) formula.
WE 4Projectile — maximum height
A ball is thrown upward and its height (m) at time t (s) is modelled by h(t) = −5t² + 20t. Find the maximum height, the time at which it occurs, and when the ball returns to the ground.
Step 1 — a = −5 < 0, so there’s a max at the vertex
Step 2 — time of maximum: t = −b/(2a)
t = −20/(−10) = 2
Step 3 — height at t = 2
h(2) = −5(4) + 20(2) = −20 + 40 = 20
Step 4 — lands when h(t) = 0
−5t² + 20t = 0
−5t(t − 4) = 0
t = 0 (launch) or t = 4 (lands)
max height 20 m at t = 2 s · ball lands at t = 4 s
classic AI SL setup: “find the maximum” ≡ “find the vertex”. The negative coefficient on t² is what guarantees a max (not a min) — check the sign first.
WE 5Find a quadratic from its features
A quadratic curve has roots at x = −2 and x = 4 and passes through the point (0, 8). Find its equation in the form y = ax² + bx + c.
Step 1 — start in factored form
y = a(x + 2)(x − 4)
Step 2 — use (0, 8) to find a
8 = a(0 + 2)(0 − 4)
8 = a × 2 × (−4)
8 = −8a ⇒ a = −1
Step 3 — expand to standard form
y = −(x + 2)(x − 4)
= −(x² − 2x − 8)
= −x² + 2x + 8
y = −x² + 2x + 8
verify: at x = −2: y = −4 − 4 + 8 = 0 ✓; at x = 4: y = −16 + 8 + 8 = 0 ✓; at x = 0: y = 8 ✓. Three independent checks — perfect.
WE 6Optimisation — fenced field
A rectangular field is enclosed on three sides using 100 m of fencing; the fourth side is a long wall. Find the dimensions that give the maximum area, and state that maximum area.
Step 1 — set up: let x be each perpendicular side, y the parallel side
fencing: 2x + y = 100
so y = 100 − 2x
Step 2 — area in terms of x alone
A(x) = x × y = x(100 − 2x)
= −2x² + 100x (quadratic in x, a = −2)
Step 3 — maximum at x = −b/(2a)
x = −100/(−4) = 25
Step 4 — other side and area
y = 100 − 2(25) = 50
A(25) = 25 × 50 = 1250
25 m × 50 m, max area = 1250 m²
a perfect AI SL optimisation: get the quantity as a quadratic in one variable, then apply the vertex formula. No calculus required.
💡 Top tips
- Use the formula booklet: the axis of symmetry x = −b/(2a) is right there — don’t try to memorise it under pressure, just look it up.
- When you have both roots, skip the formula: axis = midpoint of roots = (x1 + x2)/2. Faster and less arithmetic.
- Use the right form for the question: standard for y-intercept, factored for roots, vertex form for max/min. Mixing them is fine and often saves time.
- For max/min word problems: get the quantity (area, height, profit) as a quadratic in ONE variable. The vertex gives the optimum — no calculus needed at AI SL.
- Three points ⇒ one quadratic: a quadratic has three unknowns (a, b, c) so three independent pieces of info pin it down (e.g. two roots + one point, or vertex + one point).
⚠ Common mistakes
- Sign error in −b/(2a): there’s a minus sign in front of b. For y = x² − 4x + 3, b = −4, so axis is x = −(−4)/(2·1) = 2, NOT −2.
- Forgetting to substitute back for the vertex y-coord: −b/(2a) only gives x. You still need to plug into the equation (or use GDC) to get y.
- Wrong shape when a < 0: negative a means downward-opening with a MAX. Don’t draw a smile ∪ for a curve that should be a frown ∩.
- Reading roots backwards from factored form: y = (x − 2)(x + 6) has roots at x = 2 and x = −6 (flip the sign inside each bracket).
- Confusing local max with global max: a quadratic has only ONE extremum, so for these it’s always global. But don’t generalise that idea to cubics or higher — those have local extrema that aren’t global.
Up next: Cubic Functions & Graphs. Cubics introduce a new shape with TWO turning points (a local max and a local min) and up to three roots. The sketching workflow is similar — identify shape, find intercepts, find turning points — but the GDC does more of the heavy lifting since cubic roots rarely factorise neatly.
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