Key Idea: Reading a graph means naming its landmarks — intercepts, turning points and asymptotes — and finding where two graphs meet. It runs through both papers: by hand on Paper 1, on the GDC for Paper 2.
🗺️ The key-features menu
| Feature | What it is | How to find it |
|---|---|---|
| x-intercepts (zeros / roots) | where the curve crosses the x-axis | set y = 0 and solve |
| y-intercept | where the curve crosses the y-axis | set x = 0 |
| Max / min (turning points) | where the curve turns | vertex form, or GDC minimum/maximum |
| Vertical asymptote | line the curve shoots to ±∞ along | set the denominator = 0 |
| Horizontal asymptote | value y approaches as x → ±∞ | see what the curve levels off to |
| Increasing / decreasing | where it goes up / down (give x-intervals) | intervals change at the turning points |
Maximum value wants a y-coordinate; maximum point wants coordinates (x, y); where the maximum occurs wants the x-coordinate. The words zero, root and x-intercept all mean the same thing — set y = 0.
Where graphs cross, that point lies on both curves, so its x makes f(x) = g(x). On Paper 1, set f(x) = g(x), move everything to one side, solve, then substitute each x back for its y. On Paper 2, graph both and use intersect. Solving f(x) = k is just meeting the line y = k (k = 0 gives the x-intercepts).
✏️ IB-style worked examples
IB-style question — find the intercepts
Find the intercepts of f(x) = (x − 3)(x + 5).
Step by step:
x-intercepts (zeros): set each factor to 0.
y-intercept: set x = 0.
Zeros at x = 3 and x = −5; y-intercept (0, −15).
IB-style question — state the minimum point and value
State the minimum point and minimum value of f(x) = (x − 4)² − 7.
Step by step:
Vertex form a(x − h)² + k turns at (h, k).
a = 1 > 0, so it opens up — it's a minimum.
Minimum point (4, −7); minimum value −7.
IB-style question — find where two graphs meet (Paper 1)
Find where y = x² + 2 meets y = 3x + 2, without a calculator.
Step by step:
Set the two equal.
Bring everything to one side.
Factor and solve.
Substitute each x back (use y = 3x + 2) for its y.
They meet at (0, 2) and (3, 11).
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: Solving f(x) = g(x) (or f(x) = k) gives the x-values only. The question almost always wants points, so substitute each x back to get its y. And check for every crossing — a parabola meets a line at up to two points.
Tap each card to reveal the answer.
Find the zeros of y = (x − 1)(x + 6) x = 1 and x = −6 — set each factor to 0.
Minimum point of f(x) = (x + 2)² − 9 (−2, −9) — vertex form a(x − h)² + k turns at (h, k).
Asymptotes of f(x) = 4 + 1/(x + 3) x = −3 (vertical), y = 4 (horizontal) — denominator zero, then the leftover constant.
Where does y = x² meet y = x + 6? (−2, 4) and (3, 9) — solve x² − x − 6 = 0, then find each y.
Solve f(x) = 0 in words means… find the x-intercepts (zeros / roots) — where the graph cuts the x-axis.
Parabola y = x² − 9: where is it decreasing? x < 0 — it falls left of the vertex (x = 0) and rises right of it.
Exam tips
- Match the wording: value → y-coordinate, point → (x, y), where → x-coordinate.
- Zero = root = x-intercept (set y = 0); y-intercept comes from x = 0.
- Vertical asymptote: denominator = 0. Horizontal asymptote: what y levels off to as x → ±∞.
- Intersections solve f(x) = g(x); f(x) = k meets the line y = k. Always finish with the y-coordinate.
- Paper 2: CALC menu — 2:zero, 3:minimum, 4:maximum, 5:intersect — and check for more than one crossing.