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)
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Test this topic →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.
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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.