Sketching curves from roots, symmetry and asymptotes
Show the defining features of a curve without confusing a sketch with an accurately plotted graph.
Core: linear and quadratic sketches with roots and symmetry; turning points are not required. Extended adds turning points, cubic, reciprocal and exponential sketches, and horizontal and vertical asymptotes.
Before you begin
- Find roots and substitute x = 0 for a y-intercept.
- Complete the square and interpret the graph families.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What gives a graph's y-intercept?
What you will learn
- Sketch linear and quadratic curves with intercepts and symmetry.
- Locate Extended quadratic turning points by completing the square.
- Sketch cubic, reciprocal and exponential curves with asymptotes in Extended work.
A sketch labels the features that define its shape
For y = 2x − 4, the intercepts are (0, −4) and (2, 0). Draw a straight line crossing both axes and label these values. A sketch may not use a uniform measured scale, but it must preserve the correct direction, order and relative placement of features. An accurately plotted graph instead requires scales and calculated points.
For y = x² − 2x − 3 = (x − 3)(x + 1), roots are three and −1 and the y-intercept is −3. A positive leading coefficient gives an upward-opening curve. Its symmetry line lies halfway between the roots, x = 1. Core work uses roots and symmetry; the next paragraph's turning-point coordinates are Extended.
Extended: locate the turning point exactly
Completing the square gives y = (x − 1)² − 4, so the minimum point is (1, −4). Label it with the roots and intercept when sketching. For y = −(x + 2)² + 5, the turning point is (−2, 5) and the curve opens downward. Its maximum follows because a square is non-negative.
A cubic of the form y = ax³ + b has its basic S shape shifted vertically by b. A cubic with three simple roots crosses the horizontal axis at each root. For y = x³ − x = x(x − 1)(x + 1), roots are −1, 0 and 1. Its leading coefficient is positive, so the left end goes down and the right end goes up; local turning points are explored through differentiation later.
Extended: mark lines that are approached
For y = 3/x + 2, the vertical asymptote is x = 0 and horizontal asymptote is y = 2. An asymptote is a line the curve approaches in the relevant limit; it is not another branch of the graph. Here the curve cannot cross x = 0, and no finite input gives y = 2. Other graph families can cross an asymptote, so do not use a universal rule that crossings are impossible.
For y = 2(1/2)ˣ + 1, the output decreases towards the horizontal asymptote y = 1 and its y-intercept is three. For y = 2 × 3ˣ − 1, the curve increases and approaches y = −1 far to the left. Exponential outputs before a vertical shift are positive, so adding a constant moves their horizontal limiting line to that constant.
Pause and explain
Which way does y = −x² + 2x + 3 open?
Worked example
Sketch y = x² − 2x − 3, labelling roots and y-intercept. Extended: also label the turning point.
Show the worked solution
- Factorise to (x − 3)(x + 1), giving intercepts (−1, 0) and (3, 0).
- At x = 0, y = −3; draw an upward curve symmetric about x = 1.
- Extended: y = (x − 1)² − 4 gives the minimum (1, −4).
Answer Roots −1 and 3; y-intercept −3; symmetry x = 1; Extended turning point (1, −4)
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Sketch y = x² − 4, labelling roots, y-intercept and symmetry.
Give me a hint
Use the difference of squares and the sign of x².
Compare my reasoning
- Roots are x = −2 and 2, and the y-intercept is −4.
- The positive coefficient gives an upward-opening curve.
- Its axis of symmetry is x = 0; place the intercepts accordingly.
Roots ±2; y-intercept −4; upward curve symmetric about x = 0
Look for these in your work
- Labelled the correct intercepts.
- Preserved shape and symmetry.
Extended: sketch y = 4/x − 1, identifying both asymptotes and its x-intercept.
Give me a hint
The reciprocal shift sets the horizontal asymptote.
Compare my reasoning
- The denominator excludes x = 0, giving the vertical asymptote.
- For large input magnitude 4/x approaches zero, so y approaches −1.
- Set 4/x − 1 = 0 to obtain x = 4; the x-intercept is (4, 0).
Asymptotes x = 0 and y = −1; x-intercept (4, 0)
Look for these in your work
- Kept the reciprocal branches separate.
- Distinguished an intercept from an asymptote.
Extended: a cooling model is T = 20 + 60(1/2)ᵗ, for t ≥ 0. Describe a sketch, its starting value and its long-term level.
Give me a hint
The exponential is positive and decreases with time.
Compare my reasoning
- At t = 0, T = 20 + 60 = 80.
- At t = 1 and 2, T is 50 and 35, showing decreasing temperature.
- The graph approaches the horizontal level T = 20 from above; it never reaches it at a finite t in this model.
Starts at 80; decreases towards horizontal asymptote T = 20
Look for these in your work
- Evaluated the initial exponential correctly.
- Interpreted the limiting level within the stated domain.
Common mistakes
- Putting the turning point at the bracket's written sign.
- Drawing an asymptote as part of the curve.
- Calling a rough sketch an accurately scaled graph.
What can you explain now?
Sketch features of y = −x² + 9: roots, y-intercept and opening direction.
Compare with the explanation
Roots −3 and 3; y-intercept 9; opens downward
Solve −x² + 9 = 0 for roots, substitute x = 0 for the intercept and read the negative leading coefficient.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Sketch a quadratic using only its factored form and coefficient sign. Extended: add a shifted reciprocal and an exponential decay sketch with labelled asymptotes.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources