01 · Intercepts
x = t² − 4, y = t, with t real. Find all axis crossings.
Hint
Set each coordinate to zero separately.
Worked solution
x = 0 at t = ±2 gives (0, −2), (0, 2). y = 0 at t = 0 gives (−4, 0).
Understand · explore · practise
Sketch parametric curves using domains, intercepts, symmetry, direction and limiting behaviour. Explore a self-intersecting loop and separate rational branches, with original worked practice.
Before you startParametric coordinates, elimination, graph shapes and limits
01 / Build a sketch from justified features
Begin with the valid parameter domain. Find intercepts and endpoints, check symmetry, then use a table in increasing parameter order to connect the features. Eliminate the parameter when it helps identify the underlying curve.
Domain → key points → symmetry → tracing order → limiting behaviour
A loop or an undefined parameter can make a simple left-to-right graphing approach fail. Follow the parameter itself in the model.
x = t² − 1; y = t³ − t; −2 ≤ t ≤ 2.
t = −2 gives (3, −6).
The origin occurs at t = −1 and t = 1. The path between them is a loop.
Grey shows the displayed domain; blue follows the parameter up to your selection. The reciprocal graph continues beyond the frame near its pole. Separate branches are never joined across an undefined parameter.
02 / Find intercepts from the appropriate coordinate rule
x = 0 gives t = −1 or t = 1
Both give y = 0, so the y-axis is crossed at one point: (0, 0).
y = t(t − 1)(t + 1) = 0
The parameter values are −1, 0 and 1.
At t = 0, x = −1
The x-axis crossings are (−1, 0) and (0, 0).
Three parameter values give only two distinct x-axis points
Repeated parameter roots must be converted to positions before counting points.
An intercept parameter must also belong to the stated domain. Algebraic roots outside it do not contribute.
03 / Include values between repeated points
For the same curve, use values between t = −1 and t = 1 as well as the integer values.
t = −2 → (3, −6)
t = −1 → (0, 0)
t = −1/2 → (−3/4, 3/8)
t = 0 → (−1, 0)
t = 1/2 → (−3/4, −3/8)
t = 1 → (0, 0)
t = 2 → (3, 6)
As t increases, the curve approaches the origin along the lower-right branch, travels around the upper and then lower sides of a loop to the left, returns to the origin, and leaves along the upper-right branch.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Test symmetry using the parameter rules
x(−t) = t² − 1 = x(t)
The horizontal coordinate is unchanged.
y(−t) = −t³ + t = −y(t)
The vertical coordinate changes sign.
Every (x, y) has a matching (x, −y)
Both parameter values are allowed.
This conclusion can fail after restricting the domain. For example, keeping only t ≥ 0 removes the paired negative parameters, so symmetry of the formulas alone no longer proves symmetry of the restricted locus.
Same x, opposite y → x-axis reflection.
Opposite x, same y → y-axis reflection.
Opposite x and y → half-turn about the origin.
05 / Justify the loop algebraically
y = t(t² − 1) = tx
Factor the y-rule.
t² = x + 1
Use the x-rule.
y² = x²(x + 1), x ≥ −1
Square y = tx and substitute.
For −1 < x < 0, there are two opposite non-zero y-values
These form the two sides of the loop. They meet at (−1, 0) and (0, 0).
The origin is a self-intersection: t = −1 and t = 1 reach it by different parts of the path. The segment between those parameter values is not retraced; it goes around a loop. For x > 0, two branches extend to the right.
06 / Show only the permitted part of a curve
−1 ≤ t ≤ 1
Only the closed loop is traced.
0 ≤ t ≤ 1
Only the lower half of the loop, from (−1, 0) to (0, 0), is traced.
−2 ≤ t ≤ 2
The loop and finite pieces of both right-hand branches are included.
−1 < t < 1
The origin is excluded, but (−1, 0) remains included at t = 0.
Parameter endpoints and geometric endpoints are not interchangeable. Two ends of a parameter interval may land on the same point.
07 / Split a curve at undefined parameters
t ≠ 1
The x-rule has a zero denominator there.
t = y − 2, so x = 1/(y − 3)
Substitute using the linear y-rule.
y = 3 + 1/x, x ≠ 0
This is a translated reciprocal curve.
t < 1 gives x < 0, y < 3; t > 1 gives x > 0, y > 3
The two pieces lie on separate branches.
A plotting program that connects adjacent sample points across t = 1 can create a false line between the branches. Keep those pieces separate.
08 / Translate parameter limits into coordinate limits
As t → 1⁻, x → −∞ and y → 3⁻
This approaches the horizontal line y = 3 on the left branch.
As t → 1⁺, x → +∞ and y → 3⁺
The right branch approaches the same horizontal line.
As t → +∞, x → 0⁺ and y → +∞
This approaches the vertical line x = 0.
As t → −∞, x → 0⁻ and y → −∞
This approaches x = 0 from the other side.
Neither asymptote is part of this locus: x can never equal zero, and y = 3 would require the excluded parameter t = 1.
09 / Mark the direction on each valid branch
On the reciprocal example, as t increases below 1, x decreases from near 0⁻ towards −∞ while y increases towards 3. Above 1, x decreases from +∞ towards 0⁺ while y increases without bound.
t = 0 gives (−1, 2); t = 1/2 gives (−2, 5/2)
The left branch moves up and left.
t = 2 gives (1, 4); t = 3 gives (1/2, 5)
The right branch also moves up and left.
If t represents time, there is no defined position at t = 1 in this model. Do not describe the branches as a continuous physical jump through infinity.
10 / Audit a finished sketch
If a feature is inferred from a few samples only, treat it as a conjecture until the algebra or further analysis supports it.
11 / Your turn
x = t² − 4, y = t, with t real. Find all axis crossings.
Set each coordinate to zero separately.
x = 0 at t = ±2 gives (0, −2), (0, 2). y = 0 at t = 0 gives (−4, 0).
For the preceding curve, compare t with −t.
The square stays the same; t changes sign.
x(−t) = x(t), y(−t) = −y(t), so the locus has x-axis symmetry.
x = t² − 4, y = t for −2 ≤ t ≤ 2. Describe the path.
Track the points at −2, 0 and 2.
From (0, −2) it moves up and left to (−4, 0), then up and right to (0, 2). It traces x = y² − 4 with −2 ≤ y ≤ 2.
For x = t² − 1, y = t³ − t, how many distinct x-axis crossings are there?
Convert t = −1, 0, 1 into coordinate pairs.
Two: (0, 0) and (−1, 0). The origin is reached twice.
Why is the table containing only t = −1 and t = 1 insufficient for the loop example?
Both give the origin.
It hides the non-constant path between them. For instance t = 0 gives (−1, 0), and t = ±1/2 give opposite non-zero y-values. The point travels around a loop.
Describe the loop example on −1 ≤ t ≤ 0.
Here t is non-positive and x = t² − 1 is non-positive.
It traces the upper half of the loop from (0, 0) to (−1, 0), since y = tx ≥ 0. Both endpoints are included.
For the loop example on −1 < t < 1, is the origin included? Is (−1, 0) included?
Find every parameter for each point.
The origin needs t = ±1, so it is excluded. (−1, 0) is included at t = 0.
x = 1/(t + 2), y = t − 1, t ≠ −2. Eliminate t and identify the asymptotes.
t = y + 1.
y = 1/x − 3, x ≠ 0. Asymptotes x = 0 and y = −3.
In the preceding example, what happens as t → −2⁺?
t + 2 approaches zero through positive values.
x → +∞ and y → −3⁺. This is approach to the horizontal asymptote y = −3, not to x = −2.
Why must you not connect sample points on opposite sides of t = −2 in that example?
The original x-coordinate is undefined at −2.
The two samples belong to separate branches; the curve does not contain the straight connecting segment. Split the plot at the pole.
x = t², y = t, but only 0 ≤ t ≤ 2. Is the locus symmetric about the x-axis?
Negative t-values are not permitted.
No. The locus is y = √x on 0 ≤ x ≤ 4; reflected negative-y points are missing except the origin.
Compare x = t², y = 2t² for −1 ≤ t ≤ 1 with the loop model.
Eliminate t and follow from −1 through 0 to 1.
The first follows y = 2x from (1, 2) to (0, 0) and back along the same segment. That is retracing, not a loop enclosing a region.
x = t − 1, y = t² for 0 < t ≤ 2. Give the endpoints and their inclusion.
Substitute the limiting parameter values.
(−1, 0) is excluded and (1, 4) is included. The locus is y = (x + 1)² with −1 < x ≤ 1.
A sketch of x = 1/(t − 1), y = t + 2 includes (0, 3). Explain why it is wrong.
Check both original coordinate rules.
x = 0 is impossible for a finite valid t. y = 3 would require t = 1, where x is undefined. (0, 3) is the intersection of the asymptotes, not a point on the curve.
12 / Recap
Section 1 of 12 · Build a sketch from justified features