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Compare functions, shade regions that satisfy inequalities, distinguish solid and dashed boundaries, and find included vertices and areas.
Before you startGraphs, simultaneous equations and inequalities
01 / Compare graphs
To solve f(x) < g(x), find where the graph of f lies below the graph of g. Compare the two outputs at the same x.
f(x) = x² − 2, g(x) = 2x + 1
f(x) = g(x) ⇒ (x − 3)(x + 1) = 0
The crossings are (−1, −1) and (3, 7). Between x = −1 and x = 3, f is lower, so f(x) < g(x) exactly when −1 < x < 3. Outside this interval f is higher.
Algebra gives the same result: f(x) − g(x) < 0. A graph helps interpret the sign, while algebra locates the boundaries exactly.
For reciprocal functions, include denominator exclusions and breaks in the graph. A curve can move from positive to negative across an asymptote without crossing zero.
Blue: f(x) = x² − 2. Gold: g(x) = 2x + 1.
Between x = −1 and x = 3, the blue curve is below the gold line. Outside that interval, it is above.
02 / Points in a region
f(x) < g(x) is a condition on x. By contrast, y < f(x) describes a two-dimensional set of points (x, y) below a curve.
y < f(x): below the curve
y > f(x): above the curve
For a vertical boundary x = a, x < a means left and x > a means right. These conditions do not depend on y.
To decide which side of a boundary satisfies a rearranged inequality, test a point that is not on the boundary. The origin is convenient when it does not lie on the line.
On these pages the shading is the region that satisfies the conditions. Some questions instead shade the rejected side, so read the convention before interpreting a diagram.
2x + y > 4
First draw the boundary 2x + y = 4.
Test (0, 0): 0 > 4 is false
Shade the side away from the origin.
y > 4 − 2x
The rearranged form confirms: shade above the line.
Test (0, 5): 5 > 4 is true
The line itself is excluded because equality is not allowed.
03 / Boundaries
If rearranging involves dividing by a negative coefficient of y, reverse the inequality before choosing above or below.
−2y > x + 4
y < −x/2 − 2
This region is below a dashed line. A vertical condition such as x ≥ 0 uses a solid boundary along the y-axis and keeps the right-hand side.
y ≥ x² − 1 and y < x + 1
The parabola is solid; the line is dashed.
At (2, 3): 3 ≥ 3 is true
The point passes the first condition.
But 3 < 3 is false
It fails the second condition.
So (2, 3) is excluded
Being on one solid boundary does not override another strict condition.
04 / Build a region
For y ≥ x² − 1 and y < x + 1, shade above the solid parabola and below the dashed line. The required region is their overlap.
x² − 1 = x + 1
(x − 2)(x + 1) = 0
The boundary crossings are (−1, 0) and (2, 3). The region exists only for −1 < x < 2. At either end there is no vertical gap and the strict upper boundary excludes the crossing point.
Choose each condition separately, then “Both”, to build the picture. Adding x ≥ 0 keeps the right-hand part. It includes (0, −1), but excludes (0, 1).
For x² − 1 ≤ y ≤ 5 − x², the lower curve must be no higher than the upper curve:
x² − 1 ≤ 5 − x²
x² ≤ 3
−√3 ≤ x ≤ √3
Both boundaries are solid, so the meeting points (−√3, 2) and (√3, 2) are included.
The green region satisfies both: x² − 1 ≤ y < x + 1. It exists for −1 < x < 2. The two crossing points are excluded because the line is a strict boundary.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Vertices & area
For x ≥ 0, x ≤ 4, y ≥ 0 and y < x + 2, the boundary lines outline a trapezium.
Intersect its adjacent boundaries to find (0, 0), (4, 0), (4, 6) and (0, 2). The bottom two corners satisfy every condition. The top two lie on the strict boundary and are excluded.
Area = ½(2 + 6) × 4 = 16
The parallel vertical sides have lengths 2 and 6; their perpendicular separation is 4. The area is 16 square units.
Excluding an edge or a finite number of vertices does not change the area. It does change whether those individual points belong to the solution set.
A trapezium between x = 0, x = 4, y = 0 and the dashed line y = x + 2. Lower corners (0,0), (4,0) are included; upper corners (0,2), (4,6) are excluded.
06 / Your turn
For a region, give conditions on coordinate pairs. For a comparison of functions, give the allowed x values. Keep those two types of answer distinct.
For f(x) = x² − 4 and g(x) = x + 2, find the intersections and solve f(x) ≤ g(x).
Set the outputs equal to find the critical x values, then inspect which curve is lower.
x² − x − 6 = 0
(x − 3)(x + 2) = 0
Intersections: (−2, 0), (3, 5). The solution to f(x) ≤ g(x) is −2 ≤ x ≤ 3, including both crossings.
Describe the region 2x − 3y ≥ 6 and state the boundary style.
Isolate y, reversing the sign when dividing by −3.
−3y ≥ 6 − 2x
y ≤ (2/3)x − 2
Shade below a solid line. Equality is included.
Which of (1, 0), (2, 3) and (0, −2) satisfy both y ≥ x² − 1 and y < x + 1?
Each point must pass both tests.
(1, 0) passes: 0 ≥ 0 and 0 < 2. (2, 3) fails 3 < 3. (0, −2) fails −2 ≥ −1. Only (1, 0) belongs to the region.
For x² ≤ y ≤ 4 − x², find the possible x values and the two boundary meeting points.
The lower boundary cannot exceed the upper one.
x² ≤ 4 − x²
x² ≤ 2
−√2 ≤ x ≤ √2
The meeting points are (−√2, 2) and (√2, 2), and both are included.
The shaded triangle has vertices (−1, −1), (3, −1), (−1, 3). The vertical and horizontal boundaries are solid; the sloping boundary is dashed. Only (−1, −1) is drawn filled.
Write inequalities for the shaded triangle. State which vertices are included and find its area.
The region lies right of the vertical boundary, above the horizontal boundary and below the dashed sloping boundary.
x ≥ −1, y ≥ −1, x + y < 2
Only (−1, −1) is included. The other two vertices lie on the dashed edge. The perpendicular legs each have length 4, so the area is ½ × 4 × 4 = 8 square units.
Describe the set satisfying y > x² + 2 and y ≤ 1.
A real square is non-negative, so compare the lowest possible upper/lower requirements.
There is no region. The first inequality requires y > 2 or higher, while the second requires y ≤ 1. No coordinate pair can satisfy both.
07 / Recap
Section 1 of 7 · Compare graphs