An intersection lies on both graphs: Where two graphs cross, that point is on both curves. So its x-value makes f(x) = g(x), and its y-value is the shared output. Finding intersections = solving f(x) = g(x).
IB-style question — check a meeting point
Verify that (2, 5) lies on both y = x² + 1 and y = 2x + 1.
Step by step
- Put x = 2 into the first.
- Put x = 2 into the second.
Final answer
Both give y = 5, so (2, 5) is a common point — an intersection.
Two outputs, one point: At an intersection the two functions agree: f(a) = g(a). That single shared value is the y-coordinate of the meeting point.
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Graph both, then 'intersect': On Paper 2, type both functions into the GDC, graph them, and use the intersect tool to read each meeting point's coordinates. Set a window that shows all the crossings first.
Watch it build: graph y = x² + 1 and y = 2x + 1, then the GDC's intersect tool pins the two meeting points (0, 1) and (2, 5).
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Set them equal, move to one side, solve: On Paper 1, find intersections algebraically: set f(x) = g(x), bring everything to one side, and solve. Then put each x back into either function for the y-coordinate.
IB-style question — line meets parabola
Find where y = x² + 1 meets y = 2x + 1.
Step by step
- Set the two equal.
- Bring to one side.
- Factor and solve.
- Find each y (use y = 2x + 1).
Final answer
They meet at (0, 1) and (2, 5) — matching the GDC.
Don't forget the y-coordinates: Solving gives the x-values. The question usually wants points — substitute each x back to get y.
Solving f(x) = k is an intersection too: Solving f(x) = k is finding where the graph of f meets the horizontal line y = k. Setting k = 0 gives the x-intercepts (zeros).
IB-style question — meet the x-axis, then a line
For f(x) = x² − 5x + 6, find where the graph meets (a) the x-axis and (b) the line y = 2.
Step by step
- (a) Meets the x-axis: f(x) = 0.
- So the x-intercepts are…
- (b) Meets y = 2: set f(x) = 2.
- Factor and solve.
Final answer
(a) (2, 0) and (3, 0); (b) (1, 2) and (4, 2).
Solving f(x) = k graphically: draw the horizontal line y = k, find where it meets the curve, then read the x-values straight down.
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'Initial' means t = 0; 'same height' means set them equal: Modelling questions hide standard skills. Initial value = substitute t = 0. When are the two models equal? = solve A(t) = B(t) — graph both on the GDC and use intersect, reading every crossing in the given interval.
Graph both height models, then the GDC's intersect tool pins each meeting point — here the plants match height twice.
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IB-style question — two plant models
Over 0 ≤ t ≤ 12 weeks, Plant A (given fertilizer) has height A(t) = 5t + 8 cm and Plant B (no fertilizer) has height B(t) = 0.4t² + 12 cm. (a) Find the initial height of each plant. (b) Find the values of t when the two plants have the same height.
Step by step
- (a) Initial height = value at t = 0.
- (b) Same height means A(t) = B(t) — graph both and use the GDC 'intersect' tool.
- Read both crossings from the GDC (don't solve by hand on Paper 2).
Final answer
(a) Plant A: 8 cm, Plant B: 12 cm. (b) t ≈ 0.859 and t ≈ 11.6 weeks.
Don't stop at one crossing: A curve and a line can meet more than once — here they're equal twice. Scan the whole interval and report every solution the question asks for; missing the second crossing is the classic lost mark.