Domain: squash the curve flat onto the x-axis.
Every x the curve sits above or below gets covered by the shadow.
domain = [−1, 4]
left edge to right edge, brackets closed dots
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Domain and Range from a Graph
Domain is the curve's shadow on the x-axis, range is its shadow on the y-axis — and the lowest point on the curve beats any endpoint.
Read the graph
What is the range of the graphed function?
The range is the set of heights the curve reaches: lowest at the bottom of the dip, −3, highest at the right endpoint, 6 — so [−3, 6]. A is the domain, the classic swap. D starts from the left endpoint's height 1, but the curve dips well below it on the way.
Range: squash it sideways onto the y-axis.
Now it is heights that matter — and the bottom of the dip goes lower than either endpoint.
lowest height = −3, highest = 6
the dip beats the left endpoint's height of 1
range = [−3, 6]
every height in between is hit — the curve is unbroken
Half a circle. The upper half of the circle with centre (2, 1) and radius 4 starts and ends on the diameter, at height 1, and peaks at 1 + 4 = 5: range [1, 5]. Its domain is [2 − 4, 2 + 4] = [−2, 6]. The whole circle would reach down to 1 − 4 = −3.
Dots decide the brackets.
filled dot ●
point included — bracket [ ]
the curve reaches it
open circle ○
point excluded — parenthesis ( )
the curve approaches but never arrives
Both endpoints here are filled, so both intervals close: [−1, 4] and [−3, 6]. One open circle at (4, 6) would have turned them into [−1, 4) and [−3, 6).
Counting integers: [−1, 4] holds −1, 0, 1, 2, 3, 4 — that is 4 − (−1) + 1 = 6. With the open end, [−1, 4) loses the 4 and holds 5.
The two classic traps.
Trap one: answering the domain when the question says range — the pair [−1, 4] and [−3, 6] are both on the choices for exactly that reason. Trap two: building the range from endpoint heights only. Endpoints give 1 and 6, but the curve dips to −3 between them — always scan for the lowest and highest point of the whole curve.
Arrows mean "forever": write ∞.
An arrowhead says the graph keeps going past the edge of the grid. The shadow then has no end on that side, and we write ∞ (or −∞) for it.
| The graph | Domain | Range |
|---|---|---|
| starts at the filled point (1, −3), rises to the right with an arrow | [1, ∞) | [−3, ∞) |
| parabola opening down, vertex (−1, 4), arrows on both arms | (−∞, ∞) | (−∞, 4] |
| slanted line, arrows at both ends | (−∞, ∞) | (−∞, ∞) |
| horizontal line y = 2, arrows at both ends | (−∞, ∞) | just the one value 2 |
∞ always takes a round bracket — and the vertex's x is not a height
No graph ever reaches infinity, so the bracket beside ∞ is always ( or ). For the parabola, the range stops at the vertex's y-value, 4; the −1 belongs to x and has nothing to do with the range.
Gaps: two pieces, two intervals joined by ∪.
One piece runs from the filled point (−3, 0) up to an open circle at (−1, 2); a second runs from the filled point (2, 4) to the filled point (4, 5).
domain = [−3, −1) ∪ [2, 4]
no point sits above −1 ≤ x < 2
range = [0, 2) ∪ [4, 5]
no height between 2 and 4 is reached
Do not bridge the gap — but do merge overlaps
Writing [−3, 4] fills in x-values the graph never uses. On the other hand, if two pieces cover overlapping heights, their range shadows join into one interval — and an open circle is covered if another piece reaches that height.
On the exam
Expect arrows (∞), open and filled dots, gaps (∪), and the domain–range swap planted among the choices. Shadow the graph onto each axis separately; then check each bracket against its dot and make sure every ∞ has a round bracket.
Check against the equation y = (x − 1)² − 3.
vertex
(1, −3)
minimum −3 — range floor ✓
f(4)
9 − 3 = 6
range ceiling ✓
f(−1)
4 − 3 = 1
inside [−3, 6], not an extreme ✓
Now you try
Practice problem
Find the range
What is the range of the graphed function?
Squash the curve sideways onto the y-axis. The highest height is the turning point, 4, and it is on the curve, so 4 is included. The lowest height is at the right end, −5, but that circle is open, so −5 is never reached: (−5, 4]. B closes the bracket at the open circle. A takes only the left endpoint's height 0 up to the peak, but the curve keeps falling past 0 toward −5. C is the domain, [−1, 4), not the range.