Higher: connect line–quadratic intersections to difference zeros
Draw the original line and quadratic, retain paired coordinates, and link their common inputs to the zeros of the difference graph.
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
- Plot straight lines and smooth quadratics on signed Cartesian axes.
- Subtract bracketed expressions and work with fractions.
- Read coordinate pairs and distinguish exact data from graph estimates.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An original common point is (−3,8). Which point corresponds to it on d=y₂−y₁?
What you will learn
- Locate common points of a line and quadratic and report correctly paired coordinates.
- Construct d(x)=y₂(x)−y₁(x) and link its zeros to common x-values without replacing the original heights by zero.
- Distinguish crossings, one repeated contact, globally justified nonintersection and intersections hidden by a cropped view.
y₁=x+1 and y₂=x²−x−1. Shared original domain: all real inputs. Blue is the linear function; red is the nonlinear function. Keep both original coordinates at each common point.
Draw the original pair and subtract the linear function from the nonlinear function on paper. No input or solved intersection is selected for you. The two panels share the same horizontal scale; their vertical scales are independent.
Horizontal view: −2 to 4. Original-pair vertical view: −3 to 8; difference vertical view: −4 to 8. A finite crop neither proves nonintersection nor changes a valid coordinate or domain. Points outside a panel are omitted rather than moved to a border.
No input is preselected. Choose your own reading from the graph.
Read approximate common coordinates from the original pair. Their x-values match the difference zeros, while their original heights need not be zero.
A common point has two matching outputs
Draw the line y₁ and quadratic y₂ on the same axes, with different equation labels and a scale that can show their common points. At an intersection the same input x produces the same height in both rules. Read that x and its original common y together; a list of x-values alone does not give the requested coordinate pairs.
Use enough points to show the quadratic’s turn and both relevant sides, then draw a smooth curve rather than straight segments between table entries. Check the heights in the original line when exact inputs are available. If a coordinate comes only from the drawing, use an approximation sign and precision supported by the scale; exact-looking decimals do not improve a coarse graph.
Subtract the outputs on an aligned graph
Define d(x)=y₂(x)−y₁(x), subtracting the whole line expression with its signs. At a common input, y₂=y₁, so d=0. Conversely, d=0 on the common domain means those original outputs agree. Draw d on a separate aligned panel with the same horizontal scale and a clear d=0 baseline; link each root input between the two panels.
The paired points in the two panels have different meanings. An original intersection is (r,y₁(r)), whereas the corresponding difference point is (r,0). The original common height may happen to be zero, but that is not required. A positive difference puts the quadratic above the line at that input; a negative difference puts it below.
Distinguish a contact from a cropped view
Two crossings give two common points. A repeated zero can represent a touching contact where the difference reaches zero without changing sign; it still gives one coordinate pair, not two copies of a point. For example, a nonnegative square difference has only its zero-height contact. A drawing supports the shape, while an exact expression can check the classification after you attempt it.
Seeing no intersection in one finite window does not show that none exists. Change the view while keeping the same functions, coordinates and domains. A global claim needs a reason valid for every permitted input: for y₁=−1 and y₂=x²+1, the difference x²+2 is strictly positive for all real x, so no wider view can create an intersection. That is stronger evidence than an empty screen.
Sources for this lesson
Pause and explain
The difference between two original graphs is d(x)=(x−4)². What follows?
Worked example
Plot y₁=x+1 and y₂=x²−x−1 using −2≤x≤4 and an original-pair height view −3≤y≤8. Draw/read both common coordinate pairs to one decimal place. Form d(x)=y₂−y₁ and plot it in a separate aligned panel over the same x-window with −4≤d≤8 and a visible zero baseline. Link each common input to a difference zero, preserving the original paired height. Attempt the drawings before checking the solution.
Show the worked solution
- Draw the labelled line and a smooth quadratic. For y₂, points at x=−2,−1,0,1/2,1,2,3,4 have heights 5,1,−1,−5/4,−1,1,5,11; the last height lies above this initial frame, so the visible curve is clipped there rather than ended. The line’s outputs are x+1. The two common points are read approximately as (−0.7,0.3) and (2.7,3.7), to one decimal place.
- Subtract the whole line: d=(x²−x−1)−(x+1)=x²−2x−2. Its heights at x=−2,−1,0,1,2,3,4 are 6,1,−2,−3,−2,1,6. Draw this smooth quadratic with a clear d=0 baseline and the same horizontal scale as the original pair. Its zero inputs are about −0.7 and 2.7.
- Link those inputs vertically between the panels. The difference points are approximately (−0.7,0) and (2.7,0); the original common heights remain about 0.3 and 3.7. An algebraic post-check gives d=(x−1)²−3 and confirms two real zeros. It does not make the rounded graph readings exact coordinates.
Answer Original intersections, to one decimal place: approximately (−0.7,0.3) and (2.7,3.7). The difference is d=x²−2x−2 and its corresponding zero points are approximately (−0.7,0) and (2.7,0).
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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.
Plot y₁=−x+2 and y₂=x²−2x on an original-pair panel with −2≤x≤3, −3≤y≤9. Before revealing a difference trace, form d=y₂−y₁ yourself. Draw it on an aligned panel with the same x-window, −3≤d≤6 and a zero baseline. Locate both original common points and link their x-values to the difference zeros, retaining the paired original heights.
Give me a hint
Subtract the entire expression −x+2. After locating a common x, use the original line to recover its height and distinguish it from the difference height.
Compare my reasoning
- Draw the original line and smooth quadratic. At x=−2,−1,0,1,2,3 the quadratic heights are 8,3,0,−1,0,3; its turn is at (1,−1). Read a common point on either side of the turn before checking the algebra.
- The difference is (x²−2x)−(−x+2)=x²−x−2. At those same inputs its heights are 4,0,−2,−2,0,4. Plot it against the zero baseline. As a post-check, d=(x+1)(x−2), giving common inputs −1 and 2.
- At x=−1 the original line gives 3, and at x=2 it gives 0. Thus the original common points are (−1,3) and (2,0). Their difference partners are (−1,0) and (2,0). Keep each original input paired with its own height; the first common height is not zero.
Original common points: (−1,3), (2,0). Difference d=x²−x−2=(x+1)(x−2); its zero points are (−1,0), (2,0).
Look for these in your work
- Draw both original graphs and the aligned difference graph.
- Subtract the whole line correctly and locate both zero inputs.
- Give both original coordinate pairs and distinguish their heights from difference heights.
Compare y₁=3x−9/4 and y₂=x². Draw the original pair over −1≤x≤4 with −3≤y≤10, then form and draw d=y₂−y₁ in a separate aligned panel over the same horizontal window with −1≤d≤7 and a zero baseline. Use half-unit x ticks and readable quarter-unit heights near a common point. Locate all distinct common points, give their paired coordinates, and explain whether the curves cross or touch. Attempt the factors and coordinates before opening the solution.
Give me a hint
Look near where the vertical separation becomes smallest. Subtract the whole line, keep exact fractions when the graph/data support them, and check whether the difference changes sign at its zero.
Compare my reasoning
- Construct both labelled original traces and the difference panel before checking this result. Subtracting gives d=x²−3x+9/4. Its values at x=0,1,3/2,2,3 are 9/4,1/4,0,1/4,9/4, showing the smallest separation at x=3/2.
- After the attempt, completing the square gives d=(x−3/2)². This has one distinct zero input, x=3/2, and is positive on either side. Both original functions then give 9/4: (3/2)²=9/4 and 3(3/2)−9/4=9/4.
- Mark the original touching point (3/2,9/4) and link it to (3/2,0) in the difference panel. There is no sign change in the difference at that input, so the pair touches rather than crosses there. A squared factor records a repeated zero but does not produce a second distinct coordinate pair. Parts outside a frame remain parts of the functions, not endpoints.
One distinct touching intersection: (3/2,9/4). The difference is d=x²−3x+9/4=(x−3/2)², with one zero point (3/2,0), repeated algebraically.
Look for these in your work
- Attempt both graph panels before viewing the solved factorisation or coordinates.
- Keep the exact fractional common input and its correctly paired height.
- Explain repeated zero versus one touching point using the difference sign.
A narrow view −6≤x≤6 and −6≤y≤6 shows no meeting of y₁=22x−120 and y₂=x². Evaluate the claim that these functions never meet. Draw the same original pair in a useful wider view, such as 8≤x≤14 and 40≤y≤200, and draw d=y₂−y₁ in an aligned panel over 8≤x≤14 with −2≤d≤12. Locate every original common point and its corresponding difference zero. Explain what changing the viewport does and does not change.
Give me a hint
An empty window is not a global argument. Subtract the original line, find useful input values from the difference, and choose a height scale containing the original paired ordinates.
Compare my reasoning
- The initial window contains no visible intersection, but it is only a crop. Compute d=x²−22x+120. A post-attempt check factors it as (x−10)(x−12), so the candidate common inputs 10 and 12 lie outside the original horizontal view.
- In the expanded view, at x=8,10,11,12,14 the line heights are 56,100,122,144,188; the quadratic heights are 64,100,121,144,196. Draw the labelled original pair and its two common points. The difference heights are 8,0,−1,0,8, which show its two zeros clearly on the aligned panel.
- The original common points are (10,100) and (12,144), linked to (10,0) and (12,0) on d. Both were hidden by the first crop. The factorisation also shows there are no other real zero inputs for these functions. Changing the bounds changes what is visible, not the equations, domains or coordinates.
The global claim is false. Intersections are (10,100) and (12,144), with difference zeros (10,0) and (12,0) for d=(x−10)(x−12). Both points lie outside the first view.
Look for these in your work
- Distinguish no visible intersection from a justified global claim.
- Choose bounds showing both original common heights and both difference zeros.
- Keep the functions and paired coordinates unchanged when changing the view.
Common mistakes
- Subtracting only part of the line or losing its minus signs in y₂−y₁.
- Giving difference-zero heights as the original common ordinates, or counting a repeated zero as two points.
- Concluding that the functions never meet because a cropped plotting window shows no intersection.
What can you explain now?
For y₁=−1 and y₂=x²+1, can widening the view reveal a real common point? Give a reason that covers every real input.
Compare with the explanation
No. Their difference is x²+2>0 for every real x.
A square is nonnegative, so the quadratic is always at least 2 above the line. The global expression proves nonintersection; an empty view alone would not.
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Make it stick
Explain why a common input is a zero of y₂−y₁, and why its original common height is not usually zero. Contrast a repeated contact, a globally positive difference and an empty cropped view.
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.