Drawn tangents,right triangles and local gradients
Draw a local tangent and your own reading triangle,then estimate its signed coordinate gradient and interpret its units and context.
Section 4 · Functions (all14rows linked;self-check practice)
Exam scope and assessment notes
Original drawn-tangent teaching with four stages and twelve tasks. The original curve and named point are supplied givens;both tangent anchors and the third reading vertex come from actual learner input. Signed coordinate changes,calculated quotient and independent final gradient/unit/context are entered separately. Adequate alternative and reversed anchors and either right-corner orientation are supported. Pointer coordinates remain unsnapped. Original independent scales,negative/fractional gradients,cubic crossings and later meetings,reciprocal exclusions and context restrictions remain explicit. A horizontal zero-rise control is degenerate rather than a nondegenerate right triangle. A vertical line has no finite gradient. An auxiliary tangent crossing zero does not repair a reciprocal curve domain. Complete wrong attempts may study a separate explained reference without passing. Finite approximate comparisons do not approve written reasoning,tangent judgement,physical validity or formal assessment. Written reasoning,tangent judgement and practical validity require human review. This complete reviewed drawn-tangent teaching/practice bundle supplies the4Kstructural link;All14Functions source rows have reviewed structural teaching/practice links;other Mathematics B source gaps remain explicit in the current course map. Source mapping is separate from human drawing judgement and assessment approval. No4Ldifferentiation teaching is inferred from an internal exact reference. No official marks,mastery,specialist approval or formal assessment readiness. Single Higher-tier linear4MB1,entry context2026/27only. ISBN9780435044107is metadata only;no verified book body,chapter/Hub map or MathematicsA/Foundation/modular inheritance.
Before you begin
- Original sparse graph families and excluded inputs
- Independent Cartesian scales
- Straight-line gradients and signed fractions
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which points should supply the two main readings for a tangent-gradient triangle?
What you will learn
- Draw a reasonable tangent at a named point on the supplied original graph.
- Construct a visible right-angled reading triangle from two well-separated points on the actual tangent.
- Calculate signed rise divided by run using the original independent axis scales.
- Distinguish horizontal zero-rise controls,original curve domains and contextual limits.
Drawn tangents and local gradients: Worked
The original blue graph and named point are given. Place your own tangent anchors A and B and triangle corner C. Record your drawing and calculation,then a separate interpretation,before studying the explanation. Fixed comparisons do not award marks or mastery.
Draw at the named point and read your own triangle
Use the supplied original graph of y = x^2 − 1 and its named point P=(1,0). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision.
Horizontal: -4 to 4,step 1/2 dimensionless. Vertical: -12 to 12,step 3 dimensionless. A gradient uses dimensionless.
A: not plottedB: not plottedC: not plotted
Coordinate labels are rounded to four decimal places;the original plotted positions are retained.
My coordinate readings and calculation
Choose two well-separated visible anchors on your tangent. The third vertex should join a horizontal and a vertical reading leg. Keep the sign and use coordinate units rather than screen lengths.
Construction: Not recorded or changed
My separate gradient interpretation
Enter your final estimate independently. Decide whether the named mathematical point is admitted by the original model context. A tangent-line extension does not change that context.
Interpretation: Not recorded or changed
Draw at the named point and read your own triangle
Use the supplied original graph of y = −x^2 + 2x + 3 and its named point P=(2,3). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision.
Horizontal: -4 to 4,step 1 dimensionless. Vertical: -12 to 12,step 2 dimensionless. A gradient uses dimensionless.
A: not plottedB: not plottedC: not plotted
Coordinate labels are rounded to four decimal places;the original plotted positions are retained.
My coordinate readings and calculation
Choose two well-separated visible anchors on your tangent. The third vertex should join a horizontal and a vertical reading leg. Keep the sign and use coordinate units rather than screen lengths.
Construction: Not recorded or changed
My separate gradient interpretation
Enter your final estimate independently. Decide whether the named mathematical point is admitted by the original model context. A tangent-line extension does not change that context.
Interpretation: Not recorded or changed
Draw at the named point and read your own triangle
Use the supplied original graph of y = 3 and its named point P=(0,3). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This horizontal-contact control has zero rise and nonzero run. Identify its degenerate zero-rise construction explicitly;do not call it a nondegenerate right triangle. A straight-line graph is its own tangent. A constant graph has zero gradient at every named input.
Horizontal: -4 to 4,step 1/2 dimensionless. Vertical: -12 to 12,step 4 dimensionless. A gradient uses dimensionless.
A: not plottedB: not plottedC: not plotted
Coordinate labels are rounded to four decimal places;the original plotted positions are retained.
My coordinate readings and calculation
Choose two well-separated visible anchors on your tangent. The third vertex should join a horizontal and a vertical reading leg. Keep the sign and use coordinate units rather than screen lengths.
Construction: Not recorded or changed
My separate gradient interpretation
Enter your final estimate independently. Decide whether the named mathematical point is admitted by the original model context. A tangent-line extension does not change that context.
Interpretation: Not recorded or changed
Record every current tangent,triangle and calculation.
These are approximate checks on original authored exercises. Written reasoning,tangent judgement,model validity and formal assessment require human review. The horizontal control is explicitly degenerate. Teaching and practice are mapped to4K;eleven Functions source rows remain unlinked. Specialist approval and formal assessment readiness remain pending.
A local direction,not any line through two curve points
A gradient at a named point describes the graph’s local direction. Draw a reasonable tangent there. A chord through two different curve points usually gives a secant gradient instead. A straight-line graph is its own tangent;for a constant graph the gradient is zero at every input. A tangent to a cubic can cross the curve locally or meet it again elsewhere;global nonintersection is not a definition of tangent.
Construct an appropriate reading triangle
Choose two well-separated visible points on your actual drawn tangent. Add a horizontal and a vertical side using a third vertex. Read the two coordinate changes in one endpoint order;the points need not be other points on the original curve. The triangle belongs to the tangent line.
Keep the signed coordinate quotient and independent scales
Calculate gradient as signed rise divided by signed run. Each numerical axis step must be applied independently;screen lengths or the displayed angle do not determine the quotient. Reversing both endpoint orders preserves the gradient. A horizontal tangent has zero rise;its zero-height control is degenerate rather than a nondegenerate right triangle. A vertical line has no finite gradient.
Distinguish the curve,tangent and supplied context
A reciprocal curve excludes zero,but an auxiliary tangent line may pass through that input without adding it to the original curve. An invented contextual model admits only its stated inputs. A mathematical tangent outside that context is not an admitted physical rate. Numerical drawing tolerances and finite checks do not certify visual judgement,written reasoning or formal assessment readiness.
Sources for this lesson
Pause and explain
The original reciprocal curve excludes zero. Does a tangent line passing through x=0 add that input to the original curve?
Worked example
Task 1: Use the supplied original graph of y = x^2 − 1 and its named point P=(1,0). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. Task 2: Use the supplied original graph of y = −x^2 + 2x + 3 and its named point P=(2,3). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. Task 3: Use the supplied original graph of y = 3 and its named point P=(0,3). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This horizontal-contact control has zero rise and nonzero run. Identify its degenerate zero-rise construction explicitly;do not call it a nondegenerate right triangle. A straight-line graph is its own tangent. A constant graph has zero gradient at every named input.
Show the worked solution
- Task 1: Locate the given named point P=(1,0) on the original graph;use the two original numerical axis scales.
- Task 1: Draw a reasonable local tangent at P. These separate reference anchors are A=(-1,-4) and B=(3,4);other adequate anchors on a reasonable learner tangent can be valid.
- Task 1: For this reference,choose the third vertex C=(3,-4). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 1: Keep one anchor order: coordinate run = 3 − (-1) = 4;coordinate rise = 4 − (-4) = 8.
- Task 1: The reference quotient is rise/run = (8)/(4) = 2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 1: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 2: Locate the given named point P=(2,3) on the original graph;use the two original numerical axis scales.
- Task 2: Draw a reasonable local tangent at P. These separate reference anchors are A=(0,7) and B=(3,1);other adequate anchors on a reasonable learner tangent can be valid.
- Task 2: For this reference,choose the third vertex C=(3,7). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 2: Keep one anchor order: coordinate run = 3 − (0) = 3;coordinate rise = 1 − (7) = -6.
- Task 2: The reference quotient is rise/run = (-6)/(3) = -2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 2: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 3: Locate the given named point P=(0,3) on the original graph;use the two original numerical axis scales.
- Task 3: Draw a reasonable local tangent at P. These separate reference anchors are A=(-2,3) and B=(2,3);other adequate anchors on a reasonable learner tangent can be valid.
- Task 3: For this reference,choose the third vertex C=(2,3). The rise is zero,so the control is explicitly degenerate;do not claim three distinct right-triangle vertices.
- Task 3: Keep one anchor order: coordinate run = 2 − (-2) = 4;coordinate rise = 3 − (3) = 0.
- Task 3: The reference quotient is rise/run = (0)/(4) = 0 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 3: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 3: This horizontal-contact control has zero rise and nonzero run. Identify its degenerate zero-rise construction explicitly;do not call it a nondegenerate right triangle.
- Task 3: A straight-line graph is its own tangent. A constant graph has zero gradient at every named input.
Answer Task 1: Estimated local gradient about 2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 2: Estimated local gradient about -2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 3: Estimated local gradient about 0 dimensionless;original context allowed. Written reasoning and drawing judgement require human review.
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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.
Task 1: Use the supplied original graph of y = (1/2)x^3 − 2x + 1 and its named point P=(1,-1/2). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. Task 2: Use the supplied original graph of y = 2/x and its named point P=(-1,-2). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve. Task 3: Use the supplied original graph of y = (1/2)x + 1 and its named point P=(1,3/2). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. A straight-line graph is its own tangent. A constant graph has zero gradient at every named input.
Give me a hint
Draw the local direction at the named point,choose separated anchors on your own tangent,and form the actual horizontal/vertical reading legs. Use the original coordinate units and enter the final interpretation independently.
Compare my reasoning
- Task 1: Locate the given named point P=(1,-1/2) on the original graph;use the two original numerical axis scales.
- Task 1: Draw a reasonable local tangent at P. These separate reference anchors are A=(-2,1) and B=(2,-1);other adequate anchors on a reasonable learner tangent can be valid.
- Task 1: For this reference,choose the third vertex C=(2,1). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 1: Keep one anchor order: coordinate run = 2 − (-2) = 4;coordinate rise = -1 − (1) = -2.
- Task 1: The reference quotient is rise/run = (-2)/(4) = -1/2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 1: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 2: Locate the given named point P=(-1,-2) on the original graph;use the two original numerical axis scales.
- Task 2: Draw a reasonable local tangent at P. These separate reference anchors are A=(-3,2) and B=(1,-6);other adequate anchors on a reasonable learner tangent can be valid.
- Task 2: For this reference,choose the third vertex C=(1,2). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 2: Keep one anchor order: coordinate run = 1 − (-3) = 4;coordinate rise = -6 − (2) = -8.
- Task 2: The reference quotient is rise/run = (-8)/(4) = -2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 2: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 2: The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve.
- Task 3: Locate the given named point P=(1,3/2) on the original graph;use the two original numerical axis scales.
- Task 3: Draw a reasonable local tangent at P. These separate reference anchors are A=(-2,0) and B=(2,2);other adequate anchors on a reasonable learner tangent can be valid.
- Task 3: For this reference,choose the third vertex C=(2,0). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 3: Keep one anchor order: coordinate run = 2 − (-2) = 4;coordinate rise = 2 − (0) = 2.
- Task 3: The reference quotient is rise/run = (2)/(4) = 1/2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 3: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 3: A straight-line graph is its own tangent. A constant graph has zero gradient at every named input.
Task 1: Estimated local gradient about -1/2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 2: Estimated local gradient about -2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 3: Estimated local gradient about 1/2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review.
Look for these in your work
- My actual tangent is reasonable at the named original point.
- My two chosen tangent anchors and third reading vertex are retained.
- My signed rise and run use consistent endpoint order and independent coordinate scales.
- My final gradient,unit and context decision are entered separately.
- I explain the horizontal/domain/model limits and retain the need for human drawing review.
Task 1: Use the supplied original graph of y = x^3 and its named point P=(0,0). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This horizontal-contact control has zero rise and nonzero run. Identify its degenerate zero-rise construction explicitly;do not call it a nondegenerate right triangle. This tangent crosses the original cubic locally;local crossing does not prevent tangency. Task 2: Use the supplied original graph of y = 1/x^2 and its named point P=(-1,1). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve. Task 3: Use the supplied original graph of y = x^2 + 1/x + 1/x^2 and its named point P=(1,3). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve.
Give me a hint
Draw the local direction at the named point,choose separated anchors on your own tangent,and form the actual horizontal/vertical reading legs. Use the original coordinate units and enter the final interpretation independently.
Compare my reasoning
- Task 1: Locate the given named point P=(0,0) on the original graph;use the two original numerical axis scales.
- Task 1: Draw a reasonable local tangent at P. These separate reference anchors are A=(-2,0) and B=(2,0);other adequate anchors on a reasonable learner tangent can be valid.
- Task 1: For this reference,choose the third vertex C=(2,0). The rise is zero,so the control is explicitly degenerate;do not claim three distinct right-triangle vertices.
- Task 1: Keep one anchor order: coordinate run = 2 − (-2) = 4;coordinate rise = 0 − (0) = 0.
- Task 1: The reference quotient is rise/run = (0)/(4) = 0 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 1: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 1: This horizontal-contact control has zero rise and nonzero run. Identify its degenerate zero-rise construction explicitly;do not call it a nondegenerate right triangle.
- Task 1: This tangent crosses the original cubic locally;local crossing does not prevent tangency.
- Task 2: Locate the given named point P=(-1,1) on the original graph;use the two original numerical axis scales.
- Task 2: Draw a reasonable local tangent at P. These separate reference anchors are A=(-3,-3) and B=(1,5);other adequate anchors on a reasonable learner tangent can be valid.
- Task 2: For this reference,choose the third vertex C=(1,-3). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 2: Keep one anchor order: coordinate run = 1 − (-3) = 4;coordinate rise = 5 − (-3) = 8.
- Task 2: The reference quotient is rise/run = (8)/(4) = 2 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 2: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 2: The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve.
- Task 3: Locate the given named point P=(1,3) on the original graph;use the two original numerical axis scales.
- Task 3: Draw a reasonable local tangent at P. These separate reference anchors are A=(-2,6) and B=(2,2);other adequate anchors on a reasonable learner tangent can be valid.
- Task 3: For this reference,choose the third vertex C=(2,6). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 3: Keep one anchor order: coordinate run = 2 − (-2) = 4;coordinate rise = 2 − (6) = -4.
- Task 3: The reference quotient is rise/run = (-4)/(4) = -1 dimensionless. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 3: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 3: The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve.
Task 1: Estimated local gradient about 0 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 2: Estimated local gradient about 2 dimensionless;original context allowed. Written reasoning and drawing judgement require human review. Task 3: Estimated local gradient about -1 dimensionless;original context allowed. Written reasoning and drawing judgement require human review.
Look for these in your work
- My actual tangent is reasonable at the named original point.
- My two chosen tangent anchors and third reading vertex are retained.
- My signed rise and run use consistent endpoint order and independent coordinate scales.
- My final gradient,unit and context decision are entered separately.
- I explain the horizontal/domain/model limits and retain the need for human drawing review.
Task 1: Use the supplied original graph of T = −t^2 + 4t + 2 and its named point P=(3,5). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This is an invented temperature formula admitted only from0to4minutes;the tangent estimate is not a measured physical law. Task 2: Use the supplied original graph of q = x^2 + 1/x + 1/x^2 and its named point P=(-1,1). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This hypothetical score model is admitted only for positive inputs from1/2to2. The named negative mathematical point is outside that context;state that its tangent is not an admitted model rate. The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve. Task 3: Use the supplied original graph of s = (1/2)t^3 and its named point P=(2,4). Draw your own local tangent with two chosen anchors. Add your own third triangle vertex,then enter signed coordinate run,rise and their quotient. Enter a separate final gradient,unit and original-context decision. This invented distance formula is admitted only from0to3seconds. An auxiliary tangent may extend to negative reading heights without asserting a negative observed distance or validating a physical model.
Give me a hint
Draw the local direction at the named point,choose separated anchors on your own tangent,and form the actual horizontal/vertical reading legs. Use the original coordinate units and enter the final interpretation independently.
Compare my reasoning
- Task 1: Locate the given named point P=(3,5) on the original graph;use the two original numerical axis scales.
- Task 1: Draw a reasonable local tangent at P. These separate reference anchors are A=(0,11) and B=(4,3);other adequate anchors on a reasonable learner tangent can be valid.
- Task 1: For this reference,choose the third vertex C=(4,11). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 1: Keep one anchor order: coordinate run = 4 − (0) = 4;coordinate rise = 3 − (11) = -8.
- Task 1: The reference quotient is rise/run = (-8)/(4) = -2 degrees Celsius/minute. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 1: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 1: This is an invented temperature formula admitted only from0to4minutes;the tangent estimate is not a measured physical law.
- Task 2: Locate the given named point P=(-1,1) on the original graph;use the two original numerical axis scales.
- Task 2: Draw a reasonable local tangent at P. These separate reference anchors are A=(-3,3) and B=(1,-1);other adequate anchors on a reasonable learner tangent can be valid.
- Task 2: For this reference,choose the third vertex C=(1,3). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 2: Keep one anchor order: coordinate run = 1 − (-3) = 4;coordinate rise = -1 − (3) = -4.
- Task 2: The reference quotient is rise/run = (-4)/(4) = -1 score units/input unit. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 2: The named point is disallowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 2: This hypothetical score model is admitted only for positive inputs from1/2to2. The named negative mathematical point is outside that context;state that its tangent is not an admitted model rate.
- Task 2: The original curve excludes zero. Its auxiliary tangent line may cross that input without adding a point to the original curve.
- Task 3: Locate the given named point P=(2,4) on the original graph;use the two original numerical axis scales.
- Task 3: Draw a reasonable local tangent at P. These separate reference anchors are A=(1,-2) and B=(3,10);other adequate anchors on a reasonable learner tangent can be valid.
- Task 3: For this reference,choose the third vertex C=(3,-2). AC is horizontal and CB is vertical. Use a sizeable visible triangle,not the curve chord or a ratio of screen lengths.
- Task 3: Keep one anchor order: coordinate run = 3 − (1) = 2;coordinate rise = 10 − (-2) = 12.
- Task 3: The reference quotient is rise/run = (12)/(2) = 6 metres/second. Reversing both endpoints reverses both changes and preserves the quotient.
- Task 3: The named point is allowed by the supplied context. A mathematical tangent or its extension does not establish a physical model rate outside the original context.
- Task 3: This invented distance formula is admitted only from0to3seconds. An auxiliary tangent may extend to negative reading heights without asserting a negative observed distance or validating a physical model.
Task 1: Estimated local gradient about -2 degrees Celsius/minute;original context allowed. Written reasoning and drawing judgement require human review. Task 2: Estimated local gradient about -1 score units/input unit;original context disallowed. Written reasoning and drawing judgement require human review. Task 3: Estimated local gradient about 6 metres/second;original context allowed. Written reasoning and drawing judgement require human review.
Look for these in your work
- My actual tangent is reasonable at the named original point.
- My two chosen tangent anchors and third reading vertex are retained.
- My signed rise and run use consistent endpoint order and independent coordinate scales.
- My final gradient,unit and context decision are entered separately.
- I explain the horizontal/domain/model limits and retain the need for human drawing review.
Common mistakes
- Using a chord through two different curve points as if every secant were a tangent.
- Reading screen lengths without the two original coordinate scales.
- Taking the two coordinate differences in inconsistent endpoint orders.
- Supplying a correct final gradient while the actual line misses the named point.
- Treating a zero-rise horizontal control as a nondegenerate right triangle.
- Applying a reciprocal curve’s zero exclusion to its auxiliary tangent line.
- Treating a mathematical tangent outside a supplied context as an observed physical rate.
What can you explain now?
What happens to the estimated gradient when both endpoints of a reading triangle are reversed?
Compare with the explanation
The signed rise and run both reverse,so their quotient stays the same.
Keep one consistent endpoint order for both coordinate differences;the geometric line has not changed.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Close the reference. Draw a tangent and your own appropriate reading triangle. Calculate signed coordinate rise divided by run,explain why reversing both endpoint orders preserves the quotient,and distinguish the original curve,tangent extension and supplied context.
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.