Test a Point

Mathematics 10 · Term 1 · Week 7
Name: Section: Date:
Teacher: one page per learner, printed A4. Sections A to E count toward Written Works; see assessment.md §1. The test point has its own column everywhere a region is asked for — half of all single-inequality questions can be shaded correctly by guessing, so a shading with a blank beside it is not evidence of anything. In C, three of the six boundaries pass through the origin and the sheet does not say which. The answer key at the foot is screen-only.
Step 1 — the boundary Replace the sign with = and graph it. Broken for < and >. Solid for ≤ and ≥.
Step 2 — test a point Any point not on the boundary. Substitute. TRUE → shade that side. FALSE → shade the other side.

A Boundary and line style — fill every column, including the test point

InequalityBoundary Broken or solidTest point Which side
y > x² − 4
y ≤ 2x + 1
y ≥ x² − 2x − 3
x + y < 6

B IN, OUT, or ON? — for y ≥ x² − 4. Show each substitution

PointSubstitution IN / OUT / ONIs it a solution?
(0, 0)
(2, 0)
(1, −5)
(−3, 5)

C Can you use the origin? — if not, write a point you can use

  1. y > x² − 4
  2. y > −x
  3. y ≤ 2x + 1
  1. y < x²
  2. x + y ≥ 6
  3. y ≥ 3x

D Graph both. Mark and circle your test point.

y ≥ x² − 4
y < 2x + 1

E True or false?

  1. A broken boundary includes its own points.
  2. “>” always means shade above.
  3. The origin can always be used.
  4. A region has infinitely many points.

F In one sentence

  1. Why must the test point be off the boundary?
  2. What kind of thing is the solution of an inequality in two variables?

★ The challenge

y > x² − 4   and   x² − y < 4
  1. One says greater and one says less. Test (0, 0) in both and write both substitutions.
  2. Test (0, −10) in both.
  3. Same region or different? Justify from your two tests, not from the signs.
Answer key — screen only, not printed.
A · y = x² − 4, broken, (0, 0), 0 > −4 T → side with the origin · y = 2x + 1, solid, (0, 0), 0 ≤ 1 T → side with the origin · y = x² − 2x − 3, solid, (0, 0), 0 ≥ −3 T → side with the origin · y = −x + 6, broken, (0, 0), 0 < 6 T → side with the origin
B · (0, 0): 0 ≥ −4 → IN, yes · (2, 0): 0 ≥ 0 → ON, yes (≥ includes it) · (1, −5): −5 ≥ −3 false → OUT, no · (−3, 5): 5 ≥ 5 → ON, yes
C · 1 yes · 2 no, use (1, 1) · 3 yes · 4 no, use (0, 1) · 5 yes · 6 no, use (0, 1)  — items 2, 4 and 6 are the three that pass through the origin
D · solid parabola, vertex (0, −4), zeros ±2, shade the side with (0, 0) · broken line through (0, 1) and (1, 3), shade the side with (0, 0) since 0 < 1
E · false · false · false · true  |  F · 1. a point on the boundary makes both sides equal, so it cannot say which side · 2. a region of the plane
★ · 1. 0 > −4 TRUE, and 0 − 0 = 0 < 4 TRUE · 2. −10 > −4 FALSE, and 0 − (−10) = 10 < 4 FALSE · 3. the same region — both accept (0, 0) and both reject (0, −10)
B rows 2 and 4 are the point of the section. Both are exactly ON, and because the sign is ≥ both are still solutions. A learner who writes ON and then "no" has separated position from membership in the wrong direction.
C is marked on the three "no" rows only. Getting 1, 3 and 5 right is not evidence — the origin works there whether or not the learner checked.