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NotesMath AATopic 2.4GDC intersections
Back to Math AA Topics
2.4.22 min read

GDC intersections

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Where graphs meet = both true at once
  • Find them on the GDC
  • By hand: equate and solve
  • Meeting a horizontal line: f(x) = k
  • Two models meeting — the exam question
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

  1. Put x = 2 into the first.
  2. 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

  1. Set the two equal.
  2. Bring to one side.
  3. Factor and solve.
  4. 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

  1. (a) Meets the x-axis: f(x) = 0.
  2. So the x-intercepts are…
  3. (b) Meets y = 2: set f(x) = 2.
  4. 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

  1. (a) Initial height = value at t = 0.
  2. (b) Same height means A(t) = B(t) — graph both and use the GDC 'intersect' tool.
  3. 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.

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Write down the equation you would solve to find where y = 2x − 1 meets y = x + 5. [1 mark]

Related Math AA Topics

Continue learning with these related topics from the same unit:

2.1.1Equations of lines
2.1.2Parallel lines
2.1.3Perpendicular lines
2.1.4Perpendicular bisector
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