Higher: estimate curve gradients by drawing tangents
Draw a tangent first, choose a large signed rise/run triangle on that line, and estimate the local gradient using the actual axis scales.
Exam scope and assessment notes
Additional Higher only. Higher includes shared Foundation content. The Graphs pathway has nine shared lessons, two Higher line-equation lessons, four graph-family lessons, nine transformation lessons and these three graphical-method lessons. H:3.3D requires drawing a tangent and using a large signed rise/run triangle with actual scales and units; differentiating is not a substitute and section3.4 calculus is not credited. H:3.3E requires both line/quadratic and genuine line/cubic intersection lessons, all practice stages, paired original coordinates and the zeros of nonlinear-minus-linear. A zero on the difference graph does not replace an original intersection height. A repeated zero represents one distinct point. The nine Foundation and seven additional Higher Graphs source rows now have teaching and staged-practice links; these are links, not exhaustive subskill approval, mastery or examination readiness. The wider Maths A course still has unlinked requirements. Graph constructions and readings are approximate where stated; exact fixed-model calculations are separate checks. Written work is not automatically marked. Acknowledging a paper attempt or revealing a reference does not prove correct construction or award marks. No general solver, official mark scheme or validated cumulative assessment is supplied. Specialist review and formal assessment validation remain unestablished. Pearson’s chapter-level modular map and uncorrected-proof navigation stay separate evidence: Graphs9 map starts386 while the proof chapter starts385, and neighbouring calculus topics do not replace H3.3D. Final full-book/Teaching Hub bodies and pagination alignment remain unverified.
Before you begin
- Read signed Cartesian scales and coordinate units.
- Find a line gradient as vertical change divided by horizontal change.
- Draw smooth quadratic and cubic traces from suitable points.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Two points read from a drawn tangent are (2,7) and (6,−5). What is its signed gradient?
What you will learn
- Draw a tangent at a specified point on a nonlinear graph before inspecting a reference line.
- Estimate its signed gradient from two well-separated tangent points, using axis values and units.
- Distinguish a tangent from a secant, a horizontal slope from an undefined one, and a graph estimate from an exact model check.
y=x²−2. The fixed contact point is P(1, -1). Original curve domain: all real inputs.
Draw your own trial tangent through P by clicking a second point in the graph, or enter that point below. Judge the curve’s local direction. A trial line is your construction, not an automatically accepted tangent.
Grid step: 1 x horizontally and 1 y vertically. Read the two numbered scales independently; a screen angle or raw grid-count ratio is not the coordinate gradient.
Blue: original curve. Red: your trial line and reading triangle. Grey dashed: reference only after your construction. The view is x from −2 to 4, y from −5 to 15. A wider view does not extend the original model domain.
No trial tangent or triangle has been supplied. Draw first, then choose two well-separated points on your line.
Draw your own trial tangent through P, then choose two well-separated readings on that line. A secant through two curve points generally has a different gradient.
Construct the tangent at the chosen point
The gradient of a curved graph depends on where you are looking. Mark the required point P, then place a ruler through P in the direction the curve follows very close to P. Draw and extend that straight tangent before opening a reference construction. Compare how the curve approaches and leaves P locally; do not select a second distant curve point and assume its joining line is the tangent.
A line through two selected points of the curve is a secant. Its rise/run measures the average change between those points, which can differ from the curve’s local direction at P. A tangent is not defined by meeting the entire curve only once: a cubic tangent can meet it again elsewhere. Judge the direction near the specified contact, then check the drawing against the curve there.
Read a large signed triangle on your line
Choose two well-separated, readable points A and B on your drawn tangent. They need not be points of the original curve. Extend the tangent if a larger triangle makes the coordinates easier to read. From left to right, calculate run=x_B−x_A and signed rise=y_B−y_A. The estimate is rise/run; a falling tangent gives a negative rise with a positive run.
Read axis numbers rather than the screen angle or raw square counts. If one horizontal square represents 2 seconds and one vertical square represents 4 metres, a square-count ratio must be multiplied by 4/2 metres per second. Label the units as vertical units per horizontal unit. A horizontal tangent has zero rise and a nonzero run, so its gradient is zero; a vertical line has zero run and no finite gradient.
Report an estimate and understand its limits
Your tangent placement, line thickness and coordinate readings affect the result. Record the two readings and the rise/run so another learner can inspect your reasoning. A result from a hand-drawn tangent should be described as about or approximately a value, using precision the graph supports. A reference triangle is a comparison after your attempt, not proof that your drawing must reproduce exactly the same readings.
The examples here use curves whose local tangents have finite slopes. Drawing the tangent to y=x² at the origin gives the horizontal line y=0 and gradient zero. An extended tangent can also leave a model’s plotted height range; that straight-line extension helps measure its slope and is not a claim about the original curve outside its stated model domain. An algebraic or differentiation check, if studied separately, cannot replace this graphical construction.
Sources for this lesson
Pause and explain
A tangent rises 3 vertical squares over 2 horizontal squares. Each vertical square is 5 metres and each horizontal square is 3 seconds. Which estimate has the correct units?
Worked example
Use the smooth graph y=x²−2 in the view −2≤x≤4, −5≤y≤15. Mark P(1,−1), draw and extend your own tangent, and choose two well-separated points on that tangent to form a large triangle. Estimate the signed gradient before inspecting a reference line or the solution. Separately draw the secant through P and Q(3,7), calculate its gradient and compare the two methods.
Show the worked solution
- Mark P on the curve and draw a line following its local direction there. A few curve points, such as (0,−2), (1,−1), (2,2) and (3,7), help you inspect the curve, but joining P to Q constructs the secant instead. Complete your tangent and record your own two readable tangent points before using the following reference readings.
- After the construction, one reference tangent gives A(0,−3) and B(3,3). From left to right its rise is 3−(−3)=6 and its run is 3−0=3, giving a gradient estimate about 6/3=2. These are points on the tangent: they do not both have to lie on the original curve. Your drawn line may give slightly different readable values.
- Draw and label the secant through P(1,−1) and Q(3,7) separately. Its gradient is [7−(−1)]/(3−1)=8/2=4. This describes the change over that interval, whereas the tangent estimates the direction at P. Give the tangent result as approximately 2, not as a machine-exact measurement from a freehand graph.
Answer The drawn-tangent gradient should be about 2. A post-attempt reference triangle has signed rise 6 and run 3. The P-to-Q secant gradient is 4, which measures a different change.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Use y=4−x² in the view −2≤x≤3, −6≤y≤6. Mark (1,3), draw your own tangent and extend it before viewing a reference line. Choose two widely separated tangent points, show a large signed triangle and estimate the gradient. Then draw the secant from (1,3) to the second curve point (2,0), and explain why its gradient is a different measurement.
Give me a hint
Anchor the line at the specified point and match the curve’s direction nearby. Read two points on that drawn line from left to right, keeping the sign of the vertical change.
Compare my reasoning
- Draw the curve and tangent at (1,3) first; record your chosen tangent points and their actual axis readings. The curve falls there, so the tangent should have a negative gradient. Do not use the secant to (2,0) as the tangent construction.
- After your attempt, a reference tangent passes through A(0,5) and B(3,−1). Its signed rise is −1−5=−6 and its run is 3, giving an estimate about −2. Use your own readable points if they lie on your drawn tangent; the large triangle makes the direction and quotient easier to inspect.
- The separately labelled secant gives (0−3)/(2−1)=−3. It uses two curve points over an interval, while the tangent estimates the local direction at (1,3). Report the tangent estimate approximately, with the triangle shown.
Tangent gradient about −2; one reference triangle gives rise −6, run 3. The secant gradient is −3.
Look for these in your work
- Draw the tangent before viewing the reference and select well-separated points on it.
- Show signed rise/run from actual axis values.
- Distinguish the approximate local tangent result from the secant gradient.
Use the genuine cubic y=x³ in the view −2.2≤x≤2.5, −10≤y≤12. At (−1,−1), draw and extend your own tangent before opening a reference line, triangle or answer. Choose a large triangle on your line and estimate the signed gradient. The same tangent later meets the curve at (2,8); draw that later meeting too and explain why it does not invalidate the local tangent.
Give me a hint
Judge the curve’s direction near the specified point, then read two widely separated points on your own tangent. A later meeting elsewhere does not define the local direction at the contact.
Compare my reasoning
- Draw the cubic and mark (−1,−1). Complete your tangent locally, extend it and record a large rise/run triangle before checking any reference. The curve is rising near the contact, so the estimate should have a positive sign.
- After your attempt, a reference tangent has points A(−2,−4) and B(1,5). Its rise is 5−(−4)=9 over run 1−(−2)=3, giving gradient about 3. Both readings are from the tangent; neither needs to be another point on the cubic.
- The reference line also passes through (2,8), which is on y=x³. This distant meeting does not alter the line’s local direction at (−1,−1). Keep the original contact and later meeting visible and separately labelled. The expected graphical result is about 3; the construction and readings are the evidence for that estimate.
Gradient about 3; a reference triangle gives signed rise 9 and run 3. A tangent can meet the cubic again at (2,8) while remaining tangent at (−1,−1).
Look for these in your work
- Construct the tangent and a large triangle before inspecting the reference.
- Keep a positive signed rise/run and report an approximate gradient.
- Show the later meeting without defining tangency as a global one-meeting rule.
An idealised height model is h=t²/2 for 0≤t≤6 seconds, with height in metres. Use a graph with t from 0 to 6 and h from −8 to 20, labelled t (s) and h (m). Each horizontal square represents 2 seconds; each vertical square represents 4 metres. Draw the tangent at (4,8) before seeing a reference construction, extend it and use a large triangle to estimate its gradient. Explain why a raw square-count ratio is insufficient and why a negative-height point on the tangent is not a height prediction by the curve.
Give me a hint
Convert both sides of your triangle to metres and seconds. The tangent may be extended for a readable triangle even though the original height curve is used only on its stated time interval.
Compare my reasoning
- Draw the model curve only for 0≤t≤6, mark (4,8), then draw and extend its local tangent before using the reference. Some curve points are (0,0), (2,2), (4,8), (6,18). They help locate the curve but do not replace the tangent construction.
- After the attempt, a reference triangle uses A(0,−8) and B(6,16) on the tangent. Its height change is 24 m over a time change of 6 s, giving an estimate about 4 m/s. On the stated grid that is 6 vertical squares over 3 horizontal squares: the raw ratio 2 must be multiplied by (4 m)/(2 s) to obtain 4 m/s.
- Keep the result approximate and show the actual metre/second readings. The tangent’s negative-height extension at t=0 is part of a straight-line slope measurement, not the original model’s height there: the curve gives h(0)=0. Do not extend the curve’s permitted time domain or treat the tangent extension as a new physical model.
Tangent gradient about 4 m/s. The reference triangle gives 24 m/6 s=4 m/s; the raw square ratio 6/3=2 is not yet the physical gradient. The tangent extension is not a height prediction by h=t²/2.
Look for these in your work
- Draw the tangent first and use a visibly large triangle on it.
- Convert both axis scales and state the result in m/s.
- Keep the height model’s domain and distinguish its curve from the tangent extension.
Common mistakes
- Choosing two curve points for rise/run when the task asks for a tangent at one point.
- Using a tiny triangle, ignoring the gradient sign, or replacing axis units by raw square counts.
- Treating a reference result as a substitute for drawing, or claiming that a tangent can never meet the curve elsewhere.
What can you explain now?
Draw the tangent to y=x² at (0,0). Is its gradient zero or undefined, and why?
Compare with the explanation
It is the horizontal line y=0, with gradient 0.
Choose two different points on that tangent: the rise is 0 and the run is nonzero. Undefined finite slope would come from a zero run, as for a vertical line.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Describe how to draw a tangent, select its two reading points and report a signed gradient with axis units. Explain one reason a secant or screen angle can give the wrong result.
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.