Sketching Curves: Key Features Without Plotting Every Point
A sketch is different from a plot — instead of calculating dozens of points, you identify the handful of features that define the shape (intercepts, symmetry, turning points, asymptotes) and draw a representative curve straight from those. Core sketches linear and quadratic graphs, and does not need turning points. Extended adds cubic, reciprocal, and exponential sketches, and does require turning points — tagged throughout.
1. Sketching vs Plotting
A plot (from the previous note) uses a table of values and precise coordinates, joined carefully. A sketch is faster and less exact — it should still be recognisably the right shape, correctly labelled with intercepts and any other required features, but doesn’t need graph paper or a ruler for the curve itself (a ruler is still needed for any straight lines or axes).
2. Sketching Linear Graphs
For a straight line, you only need two features: the y-intercept, and either the gradient or a second point (often the x-intercept).
y-intercept: x = 0 → y = −3, so the line passes through (0, −3).
x-intercept: y = 0 → 0 = 2x − 3 → x = 1.5, so the line passes through (1.5, 0).
Gradient is positive (2), so the line rises left to right.
- State the y-intercept and x-intercept you’d use to sketch y = 3x + 6.
- State the y-intercept and x-intercept you’d use to sketch y = −x + 4.
Show answers
- y-intercept (0, 6); x-intercept: 0 = 3x+6 → x = −2, so (−2, 0)
- y-intercept (0, 4); x-intercept: 0 = −x+4 → x = 4, so (4, 0)
3. Sketching Quadratic Graphs
For Core, a quadratic sketch needs: the roots (from factorising), the y-intercept, the overall shape (U or ∩, from the sign of the x² coefficient), and the line of symmetry — which sits exactly halfway between the two roots. Turning points are not required at Core level.
Factorise to find roots: (x − 2)(x − 3) = 0 → x = 2 and x = 3
y-intercept: x = 0 → y = 6
Shape: U-shaped (positive x² coefficient)
Line of symmetry: halfway between the roots, at x = 2.5 — you don’t need the minimum y-value itself at Core level, just the position of the symmetry line.
- Find the roots and y-intercept you’d use to sketch y = x² − 7x + 10.
- Find the roots you’d use to sketch y = x² − 9, and state the line of symmetry.
Show answers
- (x−2)(x−5)=0 → roots x=2,5; y-intercept (0,10)
- (x−3)(x+3)=0 → roots x=−3,3; line of symmetry x=0
4. Finding the Turning Point by Completing the Square Extended only
Extended sketches need the exact turning point. Writing the quadratic in completed square form a(x + p)² + q gives it directly: the turning point is at (−p, q).
Roots: factorise → (x − 1)(x − 5) = 0 → x = 1 and x = 5
Complete the square: x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4
Turning point: (3, −4) (a minimum, since the x² coefficient is positive)
- Find the turning point of y = x² − 4x + 1 by completing the square.
- Find the turning point of y = x² + 2x − 5 by completing the square, and state whether it’s a maximum or minimum.
Show answers
- (x−2)²−4+1=(x−2)²−3. Turning point: (2, −3)
- (x+1)²−1−5=(x+1)²−6. Turning point: (−1, −6), a minimum (positive x² coefficient).
5. Sketching Cubic Graphs Extended only
Factorise to find the roots, check the y-intercept, and use the sign of the x³ coefficient to determine the overall direction: positive means the curve rises from bottom-left to top-right; negative means the reverse.
Factorise: x³ − 4x = x(x² − 4) = x(x − 2)(x + 2)
Roots: x = −2, 0, 2
y-intercept: 0 (matches the root at x = 0)
The x³ coefficient is positive, so the curve rises overall from bottom-left to top-right, with the characteristic “wiggle” between the roots.
Root: x³ = 8 → x = 2 (only one real root)
y-intercept: −8
Since there’s no x² or x term, this cubic has no turning points at all — it’s the graph of y = x³ shifted down by 8, a single smooth rising curve.
- Find the roots you’d use to sketch y = x³ − 9x.
- Find the root of y = x³ + 27, and describe the overall shape.
Show answers
- x(x²−9)=x(x−3)(x+3)=0 → roots x=−3, 0, 3
- x³=−27 → x=−3. Since there’s no x² or x term, it’s a smooth rising curve (y=x³ shifted up by 27), with no turning points.
6. Sketching Reciprocal Graphs Extended only
For y = m/x + n, the asymptotes shift away from the axes: the vertical asymptote stays at x = 0, but the horizontal asymptote moves to y = n (since m/x → 0 as x gets very large, leaving just y = n).
Vertical asymptote: x = 0 (the function is undefined there)
Horizontal asymptote: y = 1 (shifted up from y = 0, because of the “+1”)
x-intercept: 0 = 2/x + 1 → 2/x = −1 → x = −2
- State the horizontal and vertical asymptotes of y = 4/x − 2.
- Find the x-intercept of y = 4/x − 2.
Show answers
- Vertical asymptote x = 0; horizontal asymptote y = −2
- 0 = 4/x − 2 → 4/x = 2 → x = 2
7. Sketching Exponential Graphs Extended only
For y = mˣ + n, the horizontal asymptote is at y = n (shifted from y = 0), and the y-intercept is found by substituting x = 0.
Horizontal asymptote: y = −2 (shifted down from y = 0)
y-intercept: x = 0 → y = 3⁰ − 2 = 1 − 2 = −1
Shape: standard exponential growth curve, shifted down by 2, approaching but never reaching y = −2 as x decreases.
- State the horizontal asymptote and y-intercept of y = 2ˣ + 3.
- State the horizontal asymptote and y-intercept of y = 5ˣ − 4.
Show answers
- Asymptote y = 3; y-intercept: 2⁰+3 = 4
- Asymptote y = −4; y-intercept: 5⁰−4 = 1−4 = −3
- Drawing a quadratic sketch with the line of symmetry in the wrong place — it must sit exactly halfway between the two roots.
- Forgetting turning points are required for Extended sketches but not Core ones — check which tier a question is aimed at.
- Placing a horizontal asymptote at y = 0 out of habit, instead of checking for a “+n” or “−n” shift.
- Missing a root when factorising a cubic — always check for a common factor first (like a factor of x), then factorise what’s left.
- Drawing a cubic with only one turning point when it should have two (or vice versa) — check whether the cubic has an x² or x term at all; y = ax³ + d alone has no turning points.
8. Quick-Fire Challenge Round
- Core-level quadratic sketches need the exact turning point.
- The line of symmetry of a quadratic sits halfway between its two roots.
- A cubic graph can have at most two turning points.
- Shifting y = 1/x to y = 1/x + 3 moves the horizontal asymptote to y = 3.
Show answers
- False — Core needs roots, y-intercept, shape, and line of symmetry, but not the turning point itself.
- True.
- True. Ext.
- True. Ext.
- “y = x² − 6x + 8 has roots x = 2 and x = 4, so the line of symmetry is x = 3.” (is this actually an error?)
- “To find the turning point of y = x² + 4x + 1, I completed the square to get (x+2)² − 3, so the turning point is (2, −3).”
- “y = 2ˣ − 5 has a horizontal asymptote at y = 2.”
Show answers
- This one is actually correct — (2+4)/2 = 3, so x = 3 is indeed the line of symmetry. Not every “spot the error” has an error; always check rather than assuming.
- The turning point is (−p, q) when the completed square form is (x+p)² + q — since p = 2, the turning point is (−2, −3), not (2, −3).
- The asymptote is at y = −5 (matching the “−5” shift), not y = 2.
9. Mixed Practice — Bringing It All Together
- State the y-intercept and x-intercept for sketching y = 4x − 8.
- Find the roots and y-intercept for sketching y = x² − x − 6.
- State the line of symmetry for y = x² − x − 6.
Show answers
- y-intercept (0,−8); x-intercept (2,0)
- (x−3)(x+2)=0 → roots x=3,−2; y-intercept (0,−6)
- Halfway between −2 and 3: x = 0.5
- Find the turning point of y = x² − x − 6 by completing the square.
- Find the roots of y = x³ − 16x for sketching.
- State the asymptotes of y = 3/x + 4.
Show answers
- (x−0.5)²−0.25−6=(x−0.5)²−6.25. Turning point: (0.5, −6.25)
- x(x²−16)=x(x−4)(x+4)=0 → roots x=0,4,−4
- Vertical asymptote x=0; horizontal asymptote y=4
Practice at Home
A large set again — take it steadily.
- State the y-intercept and x-intercept for sketching y = 5x + 10.
- Find the roots and y-intercept for sketching y = x² − 8x + 12.
- State the line of symmetry for y = x² − 8x + 12.
- Find the roots for sketching y = x² − 16.
- Find the turning point of y = x² − 8x + 12 by completing the square.
- Find the turning point of y = x² + 6x + 4 by completing the square.
- Find the turning point of y = x² − 2x − 8, and state whether it’s a maximum or minimum.
- Find the roots for sketching y = x³ − x.
- Find the roots for sketching y = x³ − 25x.
- Find the root of y = x³ + 1, and describe whether it has turning points.
- State the asymptotes of y = 5/x − 3.
- Find the x-intercept of y = 5/x − 3.
- State the asymptote and y-intercept of y = 4ˣ − 1.
- A quadratic has roots x = −4 and x = 6, and passes through (0, −24). Find its equation in the form y = x² + bx + c, then find its turning point.
- Sketch (in words) how the graph of y = x³ − 3x² compares to y = x³ − 3x — specifically, describe how you’d find each one’s roots and what that tells you about how many times each crosses the x-axis.
- Explain why the graph of y = 1/x² never dips below the x-axis, using the fact that x² is always positive (for x ≠ 0).
Key Vocabulary Recap
Syllabus Reference
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