Gradients,rates and graph directions
A graph height tells you the dependent quantity;its gradient tells you how that quantity changes locally as the input increases. Calculate both and connect each number to the correct graph.
Section 4 · Functions (all14rows linked;self-check practice)
Exam scope and assessment notes
Original joint4Mteaching/practice bundle across three lessons and25cases,each with Worked,Guided,Independent and Application stages. Students enter their own graph observations,term-by-term differentiation,proposed derivative,original graph markers,complete cleared root lists,domain/context decisions,sign-chart witnesses and classifications. The global comparison retains stationary values,closed endpoints,open bounds,poles,infinities,attainment and all attaining inputs. Exact rational/surd and certified interval references support bounded self-checks;the finite sparse-curve and degree-five limits are tool bounds,not syllabus exclusions. Original punctures survive simplification and both graphs;constant and stationary-without-turn cases are explicit. Correct separate answers cannot replace incorrect actual working. Complete wrong attempts may study explained references without passing. The complete source and learner-route checks support a structural4Mlink only when all three original teaching/worked/practice records remain intact. All14Functions source rows have reviewed structural teaching/practice links;other Mathematics B source gaps remain explicit in the current course map. The wider Pearson curriculum remains incomplete. Supplied graph stimuli do not award separate4Jcurve-drawing,4Kdrawn-tangent or4Nmotion coverage. Written reasoning,graph judgment,model validity and specialist review remain pending. No official marks,mastery or formal assessment readiness. Single Higher-tier linear4MB1,entry context2026/27only. BookISBN9780435044107 is official metadata only;no verified full book body,chapter/Hub map or MathematicsA/Foundation/modular inheritance.
Before you begin
- Integer-power differentiation and derivative notation
- Original function domains and graph coordinates
- Signed exact substitution and rate units
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which substitution gives the local gradient at an admitted input?
What you will learn
- Calculate a named original function height separately from its local gradient.
- Read local direction from an original graph and reconcile it with the calculated derivative and an independently entered marker.
- Use the proposed derivative graph as a rate graph while retaining the original domain.
- Distinguish instantaneous and finite average rates with units and contextual limits.
Gradients,rates and graph directions: Worked
A graph height tells you the dependent quantity;its gradient tells you how that quantity changes locally as the input increases. Calculate both and connect each number to the correct graph.
Study the method and its limits
Keep the two calculations separate
Start with the original formula and domain. Rewrite integer powers without admitting excluded inputs. Differentiate term by term,then substitute the same named input into the original formula for height and into the derivative for gradient.
Mark the original point(input,height). A positive derivative gives a rising local direction,a negative derivative a falling local direction,and zero a horizontal tangent. A positive height or an x-axis crossing does not decide the gradient.
Read a derivative graph as a rate graph
A point on the derivative graph has ordinate equal to the gradient of the original graph at the same input. A derivative value above its horizontal axis indicates an increasing original graph;below indicates decreasing. A derivative-zero reading is a candidate for stationary behaviour,not automatically a turn.
Enter your own derivative before viewing its proposed graph. A plotted proposal is a representation of your input,not confirmation that the proposal is correct. Retain original reciprocal holes on both graphs.
Instantaneous rate,average rate and context
The derivative is an instantaneous rate,with units dependent quantity per input unit. Over two distinct inputs,the average rate is the change in the original quantity divided by the input change. These rates need not agree.
Keep mathematical-domain permission separate from the stated model interval. State when an input is mathematically evaluable but outside the context,and avoid claiming an invented model has been validated in the real world.
Current stage: 0 of 2 case attempts recorded;0 fixed comparisons agree. These counts do not award marks or mastery.
Original formula: y = x^2+4
Function height and local gradient answer different questions.
No extra contextual interval is imposed. Retain the original mathematical domain. Given sample inputs: -1. Rate unit: dimensionless.
- Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation.
- For each named input,enter original-domain and context decisions,original height,derivative value and local direction.
- Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Read the blue original graph first and enter your visual predictions separately from your calculations. Then reconcile them. Use explicit * and ^ in expressions. For example,3*x^2-2/x^3. Use exact surds where taught;for graphical fifth-degree roots,enter decimal readings. Enter undefined for an excluded point. Lower/upper attaining inputs use all,none,or a semicolon-separated list. The blue graph supplies the original function;it does not count as your own curve-drawing work.
My actual working
My term-by-term differentiation
My named-input and graph readings
Original graph and my proposed derivative
Blue is the supplied original graph. Orange shows your entered derivative as a proposal;it is not a correctness signal. Mark the original function points yourself. The drawing window does not restrict the mathematical domain.
Choose a row,then click the graph or use arrow keys to move its point by0.1. Hold Shift for0.01 steps. You can also enter both marker coordinates in the table. Points outside the window stay in your working and are listed below.
- My point 1: Not yet placed with two finite coordinates.
Enter a complete derivative expression to view your proposed rate graph.
Working: Not recorded or changed
My separate final answers
Enter these separately from your working. A correct answer cannot replace a missing or incorrect calculation,marker,root list,sign chart or boundary comparison.
My separate answers: My named-input and graph readings
Separate answers: Not recorded or changed
Record your current calculation,graph and interpretation working for this case.
What these comparisons cover
- The original function domain survives simplification; context bounds and the drawing window do not change it.
- The finite exact reference supports sparse powers -2 through 3 and cleared derivative degree at most five. These are self-check bounds,not exclusions from the syllabus.
- Sturm root accounting certifies every distinct real cleared root; original-domain exclusions are applied afterwards. A few plotted samples cannot certify completeness.
- First-derivative signs certify turns; a zero derivative alone does not prove a maximum or minimum. Constants have all original inputs stationary and no isolated turn or strict extremum.
- Global attained extrema are separate from unattained finite bounds and unbounded directions. Open endpoints,poles,infinities and the original model boundaries are retained.
- Quadratic-field values use exact arithmetic; other real roots use certified rational intervals. Unresolved finite comparisons require review rather than a rounded guess.
- Mathematical self-checks do not establish specialist approval,model validity or formal assessment readiness; supplied graph stimuli do not award separate curve-drawing or tangent-drawing credit.
- Exact rational and surd entries are required where taught;graphical fifth-degree roots use a0.003 input reading tolerance or less near other roots or the pole. Their height readings use0.01 tolerance.
- Own original-graph markers are checked separately with input tolerance at most0.035 and height tolerance0.15. These finite reading tolerances do not establish mathematical equality.
- Every table is learner-entered. Correct independent answers cannot replace missing or incorrect own graph markers,root lists,sign witnesses or contextual comparison working.
- Cleared roots excluded from the original domain remain in the candidate list with undefined height and no plotted point. The all-inputs case has no invented finite root list.
- Sign-chart boundaries use r1,r2,… for your candidate rows,zero-hole for the original excluded zero,and -infinity/infinity for unbounded ends. Enter a permitted rational witness and its actual derivative value in each cell.
- Global candidate rows retain stationary inputs,original-context endpoints,one-sided hole limits and infinity behaviour. Use all or none for the appropriate attaining-input statements,and separate finite input lists with semicolons.
- Showing an explained reference after a complete attempt does not imply that the attempt passed. Written reasoning,graph judgment,model validity,specialist review and formal assessment remain outside automatic approval.
Specialist review and formal assessment readiness remain pending. The supplied graphs do not award separate curve-drawing,tangent-drawing or motion coverage.
Keep the two calculations separate
Start with the original formula and domain. Rewrite integer powers without admitting excluded inputs. Differentiate term by term,then substitute the same named input into the original formula for height and into the derivative for gradient.
Mark the original point(input,height). A positive derivative gives a rising local direction,a negative derivative a falling local direction,and zero a horizontal tangent. A positive height or an x-axis crossing does not decide the gradient.
Read a derivative graph as a rate graph
A point on the derivative graph has ordinate equal to the gradient of the original graph at the same input. A derivative value above its horizontal axis indicates an increasing original graph;below indicates decreasing. A derivative-zero reading is a candidate for stationary behaviour,not automatically a turn.
Enter your own derivative before viewing its proposed graph. A plotted proposal is a representation of your input,not confirmation that the proposal is correct. Retain original reciprocal holes on both graphs.
Instantaneous rate,average rate and context
The derivative is an instantaneous rate,with units dependent quantity per input unit. Over two distinct inputs,the average rate is the change in the original quantity divided by the input change. These rates need not agree.
Keep mathematical-domain permission separate from the stated model interval. State when an input is mathematically evaluable but outside the context,and avoid claiming an invented model has been validated in the real world.
Sources for this lesson
Pause and explain
Can a graph point have positive height and negative local gradient?
Worked example
Original y = x^2+4. Function height and local gradient answer different questions. Given input: x = -1. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working. Original y = x^2-4. A point below the axis can have a positive gradient. Given input: x = 1. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Show the worked solution
- Case 1: Differentiate y=x^2+4 to obtain dy/dx=2x;the constant4 contributes0.
- Case 1: At x=-1,the original height is5 but dy/dx=-2. Locate the original point(-1,5),then show a falling local direction.
- Case 2: For y=x^2-4,the derivative is2x.
- Case 2: At x=1,the original point is(1,-3) and the local gradient is2. Mark the point below the axis with a rising local direction.
Answer Case 1: Positive height describes position above the horizontal axis;negative gradient describes a decrease as x increases. These signs can differ. Differentiate y=x^2+4 to obtain dy/dx=2x;the constant4 contributes0. At x=-1,the original height is5 but dy/dx=-2. Locate the original point(-1,5),then show a falling local direction. The supplied curve is a stimulus;the point and local direction must be entered by the learner. Case 2: A negative function value does not imply a negative rate of change. For y=x^2-4,the derivative is2x. At x=1,the original point is(1,-3) and the local gradient is2. Mark the point below the axis with a rising local direction. A correct derivative without the correct original point does not verify graph interpretation.
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.
Original y = x^3. A horizontal tangent can occur without a turn. Given input: x = 0. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working. Original y = 1/x. An original excluded input has neither a graph point nor a derivative value. Given inputs: x = -1, 0. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Give me a hint
Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Compare my reasoning
- Case 1: For y=x^3,dy/dx=3x^2.
- Case 1: At x=0,both the function height and derivative value are0. The local tangent is horizontal.
- Case 2: For y=x^(-1),dy/dx=-x^(-2)=-1/x^2;the original domain excludes0.
- Case 2: At x=-1,the point is(-1,-1) and gradient is-1. At0,neither the original function nor its derivative has a value.
Case 1: The curve continues increasing on each side;the zero gradient alone does not establish a turning point. For y=x^3,dy/dx=3x^2. At x=0,both the function height and derivative value are0. The local tangent is horizontal. A broader graph or sign argument is required to distinguish flatness from a turn. Case 2: Leave a gap at the reciprocal pole;do not plot a fabricated point at the origin or join the branches across it. For y=x^(-1),dy/dx=-x^(-2)=-1/x^2;the original domain excludes0. At x=-1,the point is(-1,-1) and gradient is-1. At0,neither the original function nor its derivative has a value. A display window around0 does not fill the mathematical domain hole.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Original y = x^3-3*x. Retain the original graph height separately from the derivative reading. Given input: x = 2. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working. Original y = x^2+1/x+1/x^2. Combine all derivative contributions and keep the reciprocal hole. Given inputs: x = -1, 0. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Give me a hint
Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Compare my reasoning
- Case 1: For y=x^3-3x,the derivative is3x^2-3.
- Case 1: At x=2,the height is8-6=2 while the gradient is12-3=9. Mark(2,2),with local direction rising.
- Case 2: Rewrite y=x^2+x^(-1)+x^(-2). Differentiate every term to obtain2x-x^(-2)-2x^(-3).
- Case 2: At x=-1,the height is1-1+1=1 and gradient is-2-1+2=-1. At0,both values remain undefined.
Case 1: Substitute into the original formula for height and into the derivative for the rate. For y=x^3-3x,the derivative is3x^2-3. At x=2,the height is8-6=2 while the gradient is12-3=9. Mark(2,2),with local direction rising. A graph point with ordinate9 would confuse the derivative graph with the original graph. Case 2: All three derivative contributions matter,and the original reciprocal exclusion survives rewriting. Rewrite y=x^2+x^(-1)+x^(-2). Differentiate every term to obtain2x-x^(-2)-2x^(-3). At x=-1,the height is1-1+1=1 and gradient is-2-1+2=-1. At0,both values remain undefined. These named-input checks do not by themselves find every stationary point of this function.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Original T = -t^2+4*t+1. Compare two instantaneous rates with the zero average rate from time1 to3. Given inputs: t = 1, 3. Stated context [0,4]. Unit: degrees Celsius/minute. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working. Original C = q^3/3+2*q+1. A mathematical value at a negative quantity is outside this invented model. Given inputs: q = 2, -1. Stated context [0,5]. Unit: USD/unit. Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Give me a hint
Enter your rewritten original expression,one coefficient-and-power differentiation row per term,and your combined derivative with notation. For each named input,enter original-domain and context decisions,original height,derivative value and local direction. Place your own original graph point and interpret your proposed derivative graph at that input. A separate final answer does not replace this working.
Compare my reasoning
- Case 1: For T=-t^2+4t+1,dT/dt=-2t+4 degrees Celsius/minute.
- Case 1: At t=1 and3,the temperatures are both4 degrees Celsius,but the instantaneous rates are2 and-2 degrees Celsius/minute.
- Case 1: The average rate from1 to3 is(4-4)/(3-1)=0 degrees Celsius/minute. Both sample times belong to the stated context[0,4].
- Case 2: For C=q^3/3+2q+1,dC/dq=q^2+2 USD/unit.
- Case 2: At q=2,the original cost is23/3 USD and its instantaneous marginal rate is6 USD/unit.
- Case 2: At q=-1,the mathematical cost is-4/3 USD and derivative is3 USD/unit,but this quantity lies outside the model context[0,5].
Case 1: Equal endpoint temperatures give a zero average rate,while the temperature can first rise and then fall between them. For T=-t^2+4t+1,dT/dt=-2t+4 degrees Celsius/minute. At t=1 and3,the temperatures are both4 degrees Celsius,but the instantaneous rates are2 and-2 degrees Celsius/minute. The average rate from1 to3 is(4-4)/(3-1)=0 degrees Celsius/minute. Both sample times belong to the stated context[0,4]. This invented model is restricted to its stated time interval and is not an experimentally validated temperature law. Case 2: Mathematical domain admission and contextual usefulness are separate decisions. An instantaneous rate is not automatically the cost of a whole extra unit. For C=q^3/3+2q+1,dC/dq=q^2+2 USD/unit. At q=2,the original cost is23/3 USD and its instantaneous marginal rate is6 USD/unit. At q=-1,the mathematical cost is-4/3 USD and derivative is3 USD/unit,but this quantity lies outside the model context[0,5]. The invented continuous cost model does not establish a realistic pricing,production or discrete-unit rule.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Common mistakes
- Confusing the original height with its local gradient.
- Deleting a real mathematical derivative candidate merely because a model later rejects it.
- Treating zero derivative as an automatic turn or a reciprocal hole as an admitted point.
- Using a few plotted samples as a complete stationary-root proof.
- Ignoring included endpoints,open boundaries,infinity or whether a bound is attained.
- Allowing correct final answers to replace missing or wrong own graph and sign-chart working.
What can you explain now?
Can a graph point have positive height and negative local gradient?
Compare with the explanation
Yes.
Height is position on the original graph;the derivative describes the local direction and rate.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Can a graph point have positive height and negative local gradient? Yes. Height is position on the original graph;the derivative describes the local direction and rate. Does a fixed self-check establish specialist approval or formal assessment readiness? No. Written reasoning,graph judgment and model validity still need human review.
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.