Introduction to Linear Polynomials — Advanced Question Bank
A relatively difficult, source-aligned practice module covering univariate polynomials and degree, evaluation, linear polynomials, linear equations, input-output processes, linear patterns, growth and decay, relationships y=ax+b, graphical interpretation, slope, y-intercept and parallel lines.
High-Value Revision
Polynomial language
A one-variable polynomial uses one variable and non-negative integral powers. The highest power is its degree. A degree-1 polynomial is linear.
Linear pattern
For a linear pattern, consecutive values change by a constant amount. A typical rule is Tₙ=an+b.
Linear relationship
y=ax+b: a is the slope and b is the y-intercept. The line cuts the y-axis at (0,b).
Parallel-line rule
Lines with the same slope a but different y-intercepts are parallel.
Visual Learning — HTML Canvas
Only the diagrams use JavaScript. All answer buttons are native HTML <details> elements.
Section A — 30 MCQs
Most questions require calculation, interpretation or graph reasoning rather than direct recall.
Which expression is a linear polynomial in one variable?
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Answer: B
A linear polynomial in one variable has degree 1. Hence 5x − 7 is linear.
The degree of the polynomial −9 is:
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Answer: A
A non-zero constant polynomial has degree 0.
In 5y³ + y² + 2y − 1, the coefficient of y² is:
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Answer: B
The y² term is 1·y², so its coefficient is 1.
For p(x)=7−4x, p(−3) equals:
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Answer: B
p(−3)=7−4(−3)=7+12=19.
The successive values of a linear polynomial at consecutive integer inputs differ by:
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Answer: A
For p(x)=ax+b, p(n+1)−p(n)=a, which is constant.
The nth term of 1, 3, 5, 7, … is:
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Answer: B
Substituting n=1 gives 1 and the constant difference is 2, so the rule is 2n−1.
If p(x)=ax+b and p(2)=11, p(5)=23, then a equals:
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Answer: B
The output changes by 12 while x changes by 3, so a=12/3=4.
For y=−3x+5, the y-intercept is:
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Answer: C
In y=ax+b, b is the y-intercept. Here b=5.
Which equation represents a line through the origin?
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Answer: B
A line y=ax+b passes through the origin when b=0.
Which pair of lines is parallel?
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Answer: B
Parallel lines have equal slopes and different y-intercepts.
A linear growth model is most naturally represented by:
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Answer: A
The chapter identifies linear growth with a straight line having positive slope.
A linear decay model is represented by:
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Answer: B
Linear decay decreases by a fixed amount over equal intervals, giving a negative slope.
A phone costs ₹10,000 and loses ₹800 per year. Its value after t years is:
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Answer: B
Initial value is 10000 and it decreases by 800 each year.
A plant is 1.75 ft tall and grows 0.5 ft monthly. Its height after 7 months is:
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Answer: C
h=1.75+0.5(7)=1.75+3.5=5.25 ft.
For y=2x+1, which point lies on the line?
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Answer: B
For x=3, y=2(3)+1=7.
The line passing through (−1,−3), (0,0), (1,3), (3,9) is:
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Answer: B
For every listed point, y=3x.
The line through (−3,6), (0,0), (2,−4) is:
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Answer: B
Each point satisfies y=−2x.
If a>1 in y=ax with a>0, compared with y=x the line is:
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Answer: C
The chapter notes that for a>1, y=ax is steeper than y=x.
If 0<a<1 in y=ax, compared with y=x the line is:
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Answer: B
For positive a<1, the line rises more slowly and is less steep than y=x.
The lines y=2x−1, y=2x+1 and y=2x+5:
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Answer: B
All have slope 2 but different y-intercepts.
A learning platform charges according to y=ax+b. The bill is ₹400 for 10 modules and ₹500 for 14 modules. The value of a is:
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Answer: B
a=(500−400)/(14−10)=100/4=25.
Using the same data, the fixed monthly fee b is:
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Answer: B
400=25(10)+b ⇒ b=150.
The graph of p(x)=3x+2 cuts the y-axis at:
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Answer: B
At the y-axis x=0, so p(0)=2 and the point is (0,2).
If p(x)=ax+b and p(0)=−4, then:
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Answer: B
p(0)=b, therefore b=−4.
The graph of p(x) passes through (1,5) and (3,11). p(x) is:
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Answer: B
Slope=(11−5)/(3−1)=3; using 5=3(1)+b gives b=2.
For the polynomial in the previous question, the x-axis intercept is:
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Answer: A
Set y=0: 3x+2=0 ⇒ x=−2/3.
If f(x)=ax+a with a>0, every graph passes through:
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Answer: B
f(−1)=−a+a=0, so every graph passes through (−1,0).
A sequence has nth term 7n−4. The common difference is:
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Answer: C
Tₙ₊₁−Tₙ=[7(n+1)−4]−(7n−4)=7.
If y=−2x+7 and x increases by 5, the value of y changes by:
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Answer: B
Change in y=a·change in x=(−2)(5)=−10.
A line has y-intercept 4 and passes through (3,−2). Its equation is:
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Answer: B
−2=3a+4 ⇒ 3a=−6 ⇒ a=−2, so y=−2x+4.
Section B — 15 One-Mark Questions
Short but concept-focused.
State the degree of 4z−3.
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1.
Write the coefficient of x in 9−7x.
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−7.
Evaluate 5x−3 at x=−1.
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5(−1)−3=−8.
Write the y-intercept of y=−6x+11.
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11.
Write the slope of y=4x−9.
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4.
Does (2,5) lie on y=2x+1?
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Yes, because 2(2)+1=5.
Find the next term of 11, 17, 23, 29, …
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35.
Write the nth term of 4, 9, 14, 19, …
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5n−1.
A line y=ax+b passes through the origin. What is b?
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b=0.
What happens to a line y=ax+b when only b changes?
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Its slope stays the same; the line shifts parallel to itself.
Find p(0) if p(x)=−3x+8.
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8.
Find x if 2x+7=31.
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x=12.
A ₹600 balance decreases by ₹15 daily. Write the balance after x days.
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b(x)=600−15x.
Write the line through the origin with slope −3.
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y=−3x.
If two non-identical lines y=ax+b₁ and y=ax+b₂ have the same a, what is their relation?
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They are parallel.
Section C — 15 Two-Mark Questions
Two-step applications and linear-pattern reasoning.
A linear polynomial p(x)=ax+b satisfies p(2)=9 and p(5)=18. Find a.
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a=(18−9)/(5−2)=9/3=3.
Using p(2)=9 and p(5)=18, find p(x).
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From a=3, 9=3(2)+b ⇒ b=3. Hence p(x)=3x+3.
Find the 15th and 26th terms of the pattern 1,3,5,7,…
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Tₙ=2n−1. T₁₅=29 and T₂₆=51.
Which stages of the pattern 2n−1 contain 21 tiles and 47 tiles?
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2n−1=21 ⇒ n=11. 2n−1=47 ⇒ n=24.
A taxi fare is modelled by F(n)=15n−5 for n≥2. Find the distance n when the fare is ₹130.
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15n−5=130 ⇒15n=135 ⇒n=9 km.
A journey cost is C(d)=100+60d. Find d if C=₹700.
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100+60d=700 ⇒60d=600 ⇒d=10 km.
The water height is h(t)=3−0.5t. Find when the tank reaches 0.5 m.
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3−0.5t=0.5 ⇒0.5t=2.5 ⇒t=5 months.
Find the equation of a line with slope 5 and y-intercept −3.
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Using y=ax+b, y=5x−3.
Check whether (−2,7) lies on y=−3x+1.
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For x=−2, RHS=−3(−2)+1=7, so yes.
Find the fixed fee and rate per unit in y=18x+125.
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Rate per unit (slope)=₹18; fixed fee (y-intercept)=₹125.
The values of a linear pattern at n=4 and n=10 are 19 and 43. Find its common difference.
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Change in output=24 over 6 steps, so common difference=24/6=4.
A line y=ax+6 passes through (4,−2). Find a.
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−2=4a+6 ⇒4a=−8 ⇒a=−2.
If y=3x−1, compare the values of y at x=8 and x=3.
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At x=8, y=23; at x=3, y=8. Difference=15, matching slope×input change=3×5.
A rally begins with 120 members and loses 9 each hour. Write the model and find the number after 8 hours.
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R(n)=120−9n; R(8)=120−72=48.
A book has 500 pages and 20 pages are read daily. Write pages left after n days and find when 100 pages remain.
View answer / solution
P(n)=500−20n. Set 500−20n=100 ⇒n=20 days.
Section D — 15 Three-Mark Questions
Finding models, slopes, intercepts, parallel lines, growth and decay.
A gym bill is ₹800 for 10 court-hours and ₹1100 for 15 court-hours. Find the linear relationship y=ax+b.
View answer / solution
a=(1100−800)/(15−10)=300/5=60. Then 800=60(10)+b ⇒b=200. Hence y=60x+200.
Find the Celsius-Fahrenheit relation in the form C=aF+b using (F,C)=(32,0) and (212,100).
View answer / solution
a=(100−0)/(212−32)=100/180=5/9. Then 0=(5/9)(32)+b ⇒b=−160/9. Hence C=(5/9)F−160/9=(5/9)(F−32).
The graph of p(x) passes through (1,5) and (3,11). Find p(x) and both axis-intercepts.
View answer / solution
Slope a=(11−5)/(3−1)=3. Using (1,5), b=2, so p(x)=3x+2. y-axis: x=0 ⇒(0,2). x-axis: 3x+2=0 ⇒(−2/3,0).
A line has equation 2y=4x+7. Find its slope, y-intercept and the point where it cuts the y-axis.
View answer / solution
y=2x+7/2. Slope=2, y-intercept=7/2, and y-axis point=(0,7/2).
A mobile phone worth ₹10,000 depreciates by ₹800 per year. Find its value after 3 years, write the model, and find after how many years its model reaches ₹3,600.
View answer / solution
V(t)=10000−800t. V(3)=7600. Set 10000−800t=3600 ⇒800t=6400 ⇒t=8 years.
A village has population 750 and gains 50 persons per year. Write the model, find population after 6 years, and determine when it reaches 1250.
View answer / solution
P(t)=750+50t. P(6)=1050. 750+50t=1250 ⇒t=10 years.
Show that y=3x−1, y=3x+4 and y=3x−7 represent parallel lines. State their y-intercepts.
View answer / solution
All three have the same slope 3, so they are parallel. Their y-intercepts are −1, 4 and −7 respectively.
The outputs of a linear function are 17 at x=2 and −3 at x=7. Find the function.
View answer / solution
a=(−3−17)/(7−2)=−20/5=−4. Then 17=−4(2)+b ⇒b=25. Thus y=−4x+25.
A linear pattern has T₆=31 and common difference 5. Find its nth-term rule and T₂₀.
View answer / solution
Tₙ=T₆+(n−6)5=31+5n−30=5n+1. Thus T₂₀=101.
The amount A after n months follows A=500+150n. Find A after 8 months, identify the slope and intercept, and explain their meanings.
View answer / solution
A(8)=1700. Slope 150 means ₹150 added per month. Intercept 500 is the initial amount at n=0.
A line y=ax+b passes through (−2,10) and (4,−8). Find a and b.
View answer / solution
a=(−8−10)/(4−(−2))=−18/6=−3. Then 10=−3(−2)+b ⇒b=4. So y=−3x+4.
Find an equation for a line parallel to y=5x−7 and passing through (2,9).
View answer / solution
Parallel line has slope 5: y=5x+b. Using (2,9): 9=10+b ⇒b=−1. Hence y=5x−1.
A balance decreases linearly from ₹900 to ₹540 in 6 days. Find the daily decrease and a model B(t), assuming t=0 initially.
View answer / solution
Decrease=360 in 6 days, i.e. ₹60/day. Hence B(t)=900−60t.
For f(x)=2x+5 and g(x)=−x+8, find x for which f(x)=g(x), and the common value.
View answer / solution
2x+5=−x+8 ⇒3x=3 ⇒x=1. Common value=7.
A linear function has f(0)=7 and f(5)=−8. Find f(−3).
View answer / solution
Slope=(−8−7)/5=−3. Thus f(x)=−3x+7 and f(−3)=16.
Section E — 10 Four-Mark Questions
Multi-step competency and higher-order problems.
Let p(x)=ax+b and q(x)=cx+d satisfy p(0)=5, p(x)+q(x)=6x+4, and p(x)−q(x) cuts the x-axis at (3,0). Find p(x) and q(x).
View answer / solution
p(0)=b=5. From p+q=(a+c)x+(b+d)=6x+4, so a+c=6 and 5+d=4 ⇒d=−1. Since p−q=(a−c)x+(b−d) and it is 0 at x=3: 3(a−c)+(5−(−1))=0 ⇒3(a−c)+6=0 ⇒a−c=−2. Solving a+c=6 and a−c=−2 gives a=2, c=4. Hence p(x)=2x+5 and q(x)=4x−1.
A linear polynomial p(x) passes through (2,3) and (6,11). A second linear polynomial q(x) is parallel to p(x) and passes through (4,−1). Find p, q and their x-axis intercepts.
View answer / solution
For p: slope=(11−3)/(6−2)=2, so p=2x+b. 3=4+b ⇒b=−1, hence p=2x−1. q is parallel, so slope=2: q=2x+d. −1=8+d ⇒d=−9, hence q=2x−9. p cuts x-axis at x=1/2; q cuts x-axis at x=9/2.
A two-digit number has digits differing by 3. The number plus the number with digits reversed is 143. Find both possible numbers.
View answer / solution
Let tens digit be x and units digit y. Then (10x+y)+(10y+x)=143 ⇒11(x+y)=143 ⇒x+y=13. Also |x−y|=3. Solving x−y=3 gives x=8,y=5 →85; solving y−x=3 gives x=5,y=8 →58. Thus the numbers are 85 and 58.
A matchstick pattern is made from a chain of hexagons. Stage 1 uses 6 sticks; each new hexagon shares one side with the previous one. Find the nth-stage rule, sticks at stage 15, and decide whether 200 sticks can form a stage.
View answer / solution
First hexagon uses 6 sticks. Each added hexagon shares one side, so it adds 5 sticks. Thus M(n)=6+5(n−1)=5n+1. M(15)=76. For 200 sticks: 5n+1=200 ⇒5n=199 ⇒n=39.8, not an integer. Therefore 200 sticks cannot be exactly a stage.
A Kelvin-Fahrenheit relation is y=(9/5)(x−273)+32, where x is K and y is °F. Find y at 313 K, find x at 158°F, and rewrite the relation in y=ax+b form.
View answer / solution
At x=313: y=(9/5)(40)+32=72+32=104°F. For y=158: 158−32=126=(9/5)(x−273); x−273=70, so x=343 K. Expanded form: y=(9/5)x−(9/5)273+32=(9/5)x−2297/5.
A quantity is modelled by y=ax+b. At x=4, y=26; at x=10, y=−4. Find the model, identify growth/decay, find the y-intercept and determine when y=0.
View answer / solution
a=(−4−26)/(10−4)=−30/6=−5. Then 26=−20+b ⇒b=46. So y=−5x+46. Negative slope means linear decay. y-intercept=46. For y=0: −5x+46=0 ⇒x=46/5=9.2.
Three parallel service plans are y=12x+80, y=12x+140 and y=12x−40. Explain why they are parallel, identify all y-intercepts, and find the x-value where the third plan reaches y=200.
View answer / solution
All have slope 12, so they are parallel. y-intercepts are 80, 140 and −40. For 12x−40=200:12x=240⇒x=20.
A linear pattern has T₃=14 and T₁₁=46. Find its nth term, T₅₀, and the stage at which the value is 202.
View answer / solution
Difference per step=(46−14)/(11−3)=32/8=4. Let Tₙ=4n+b. 14=12+b⇒b=2, so Tₙ=4n+2. T₅₀=202. For Tₙ=202, 4n+2=202⇒n=50.
A graph passes through A(−3,11), B(1,−1). Find its equation. A second line has the same y-intercept but slope 2. Find its equation and the point where the two lines meet.
View answer / solution
First slope=(−1−11)/(1−(−3))=−12/4=−3. Using B: −1=−3(1)+b⇒b=2. First line y=−3x+2. Second line y=2x+2. Equate: −3x+2=2x+2⇒x=0, y=2. They meet at (0,2).
For f(x)=ax+a, a>0: prove all such lines pass through one fixed point; then compare the slopes for a=1/2, 1 and 3.
View answer / solution
f(x)=a(x+1). At x=−1, f(−1)=0 for every a, so all lines pass through the fixed point (−1,0). Their slopes are respectively 1/2, 1 and 3; hence the line with a=3 is steepest and a=1/2 least steep among the three.
Section F — 5 Case Studies
Context-based applications drawn from the chapter’s modelling style.
Case Study 1 — Digital Learning Platform
A digital learning platform charges a fixed monthly fee plus a fee per module. A student pays ₹400 for 10 modules and ₹500 for 14 modules. Assume y=ax+b.
- a) Find the per-module charge a.
- b) Find the fixed fee b.
- c) Write the bill model.
- d) Find the bill for 22 modules.
View case-study solutions
a) Find the per-module charge a. a=(500−400)/(14−10)=₹25 per module.
b) Find the fixed fee b. 400=25×10+b ⇒ b=₹150.
c) Write the bill model. y=25x+150.
d) Find the bill for 22 modules. y=25×22+150=₹700.
Case Study 2 — Water Level: Linear Decay
A cylindrical tank has water height 3 m initially. The height after t months is h(t)=3−0.5t.
- a) What is the monthly change in height?
- b) Find h(4).
- c) After how many months does the model reach zero?
- d) What feature of the graph signals decay?
View case-study solutions
a) What is the monthly change in height? A decrease of 0.5 m per month.
b) Find h(4). h(4)=3−2=1 m.
c) After how many months does the model reach zero? 3−0.5t=0 ⇒t=6 months.
d) What feature of the graph signals decay? Its slope is negative: −0.5.
Case Study 3 — Three Internet Plans
Three internet plans are represented by A: y=20x+150, B: y=20x+250 and C: y=30x+150, where x is data use and y is the bill.
- a) Which two plans have parallel graphs?
- b) Which two plans have the same y-intercept?
- c) Find when A and C charge the same amount.
- d) At x=10, compare B and C.
View case-study solutions
a) Which two plans have parallel graphs? A and B, because both have slope 20.
b) Which two plans have the same y-intercept? A and C, because both have y-intercept 150.
c) Find when A and C charge the same amount. 20x+150=30x+150 ⇒x=0.
d) At x=10, compare B and C. B=450 and C=450, so they are equal at x=10.
Case Study 4 — Matchstick Hexagon Pattern
A chain begins with one hexagon made from 6 matchsticks. Each new hexagon shares exactly one side with the previous hexagon.
- a) List the first five stage totals.
- b) Write the nth-stage rule.
- c) Find M(25).
- d) Can 151 sticks form a complete stage?
View case-study solutions
a) List the first five stage totals. 6, 11, 16, 21, 26.
b) Write the nth-stage rule. M(n)=5n+1.
c) Find M(25). M(25)=126.
d) Can 151 sticks form a complete stage? 5n+1=151 ⇒n=30, so yes, Stage 30.
Case Study 5 — Two Linear Polynomials
Two linear polynomials p(x)=ax+b and q(x)=cx+d satisfy p(0)=5, p(x)+q(x)=6x+4, and p(x)−q(x)=0 at x=3.
- a) Find b.
- b) Find d.
- c) Find a and c.
- d) State p(x) and q(x).
View case-study solutions
a) Find b. b=p(0)=5.
b) Find d. b+d=4 ⇒ d=−1.
c) Find a and c. a+c=6 and 3(a−c)+6=0 ⇒a−c=−2. Hence a=2,c=4.
d) State p(x) and q(x). p(x)=2x+5 and q(x)=4x−1.
Special Exemplar Challenge — 12 Questions
Non-routine questions requiring interpretation, model-building and careful use of slope/intercept ideas.
A linear polynomial p(x) has p(−2)=13 and p(4)=−5. Find p(x).
View worked solution
a=(−5−13)/(4+2)=−18/6=−3. 13=−3(−2)+b ⇒b=7. Thus p(x)=−3x+7.
A sequence is generated by a linear polynomial. Its 8th term is 31 and its 19th term is 75. Find the nth term.
View worked solution
a=(75−31)/(19−8)=44/11=4. Let Tₙ=4n+b. 31=32+b⇒b=−1. Hence Tₙ=4n−1.
Find a and b if y=ax+b passes through (−3,8) and has y-intercept −1.
View worked solution
b=−1. Then 8=−3a−1 ⇒−3a=9⇒a=−3. So y=−3x−1.
A line y=ax+5 is parallel to 4y=12x−7. Find a.
View worked solution
4y=12x−7 ⇒ y=3x−7/4, so slope=3. Parallel lines have equal slope, hence a=3.
For which value of k are y=(k−1)x+4 and y=5x−9 parallel?
View worked solution
Equal slopes: k−1=5 ⇒k=6.
The graph of y=px+q passes through (0,−4) and (5,11). Find p+q.
View worked solution
q=−4. p=(11+4)/5=3. Therefore p+q=−1.
A linear decay has value 92 at t=4 and 44 at t=10. Find the model and initial value.
View worked solution
Slope=(44−92)/6=−8. y=−8t+b. 92=−32+b⇒b=124. Model y=124−8t; initial value=124.
A pattern has constant difference −6 and its 5th term is 17. Find the first term and nth term.
View worked solution
Tₙ=T₁+(n−1)(−6). 17=T₁−24⇒T₁=41. Hence Tₙ=41−6(n−1)=47−6n.
A line cuts the y-axis at (0,7) and the x-axis at (14,0). Find its equation in y=ax+b form.
View worked solution
b=7. Using (14,0):0=14a+7⇒a=−1/2. Hence y=−x/2+7.
A linear function f satisfies f(2)+f(6)=40 and f(4)=20. What can you conclude? Describe all possible such functions.
View worked solution
For any linear f, f(2)+f(6)=2f(4) because 4 is the midpoint of 2 and 6. Since 2f(4)=40, the first condition adds no new restriction. Infinitely many lines through (4,20) work: f(x)=a(x−4)+20.
Can lines y=2x+r and y=sx−3 be parallel if the second passes through (2,5) and the first passes through (0,7)? Justify.
View worked solution
The first condition gives r=7. Parallelism requires s=2. But then the second line gives y=2(2)−3=1 at x=2, not 5. Therefore no such pair of lines exists; the conditions are inconsistent.
A line y=ax+b has a positive slope and passes through (−2,−3) and (4,9). Find a,b and explain why this is linear growth.
View worked solution
a=(9+3)/(4+2)=12/6=2; −3=2(−2)+b⇒b=1. y=2x+1. Since a=2>0, outputs increase by 2 for each unit increase in x, so it represents linear growth.
Special Olympiad / HOTS Challenge — 12 Questions
Advanced reasoning within and just beyond the chapter’s core framework, without requiring calculus or senior-class algebra.
Let p(x)=ax+b be linear and p(p(x))=4x+3 for all x, with a>0. Find a and b.
View worked solution
p(p(x))=a(ax+b)+b=a²x+b(a+1). Thus a²=4 ⇒a=2 (a>0). Then 3b=3 ⇒b=1. So p(x)=2x+1.
A linear function f satisfies f(1)=4. If f(x+2)−f(x)=6 for every x, find f(100).
View worked solution
For f(x)=ax+b, f(x+2)−f(x)=2a=6 ⇒a=3. f(1)=3+b=4⇒b=1. Hence f(100)=301.
Two linear functions p and q satisfy p(x)+q(x)=10x+6 and p(x)−q(x)=2x−8. Find p(x) and q(x).
View worked solution
Add the equations:2p=12x−2⇒p=6x−1. Subtract:2q=8x+14⇒q=4x+7.
A line passes through (2,7). When x increases by 5, y decreases by 15. Find its equation.
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Slope=−15/5=−3. Use 7=−3(2)+b⇒b=13. Equation y=−3x+13.
For what value of m do the lines y=(m²−5)x+2 and y=4x−7 become parallel? Give all real values.
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m²−5=4 ⇒m²=9 ⇒m=±3.
A family of lines is y=kx+k+2. Show all lines in the family pass through one fixed point.
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Rewrite y−2=k(x+1). At x=−1, y−2=0 for every k, so every line passes through (−1,2).
A linear function f has f(−3)=12 and f(5)=−4. Find the unique x for which f(x)=0.
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Slope=(−4−12)/(5+3)=−16/8=−2. f(x)=−2x+b; 12=6+b⇒b=6. Set −2x+6=0⇒x=3.
A pattern follows Tₙ=an+b. If T₂+T₈=50 and T₅=25, what does this tell you about a and b?
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For a linear pattern, T₂+T₈=2T₅ automatically because 5 is the midpoint of 2 and 8. Thus the first condition gives no new information beyond T₅=25. Any a,b with 5a+b=25 works; infinitely many patterns.
Find all linear functions f(x)=ax+b such that f(0)=2 and f(f(0))=8.
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b=2. f(f(0))=f(2)=2a+2=8⇒a=3. Thus uniquely f(x)=3x+2.
Three distinct parallel lines have y-intercepts in arithmetic progression 2, k, 14. Find k. If their common slope is −3, write all three equations.
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Middle term k=(2+14)/2=8. Equations: y=−3x+2, y=−3x+8, y=−3x+14.
A line y=ax+b passes through (1,1). Its x-intercept is twice its y-intercept in numerical value, and both intercepts are positive. Find the line.
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y-intercept=b>0. x-intercept=−b/a=2b ⇒−1/a=2 ⇒a=−1/2. Using (1,1):1=−1/2+b⇒b=3/2. Thus y=−x/2+3/2.
Let p(x)=3x+r and q(x)=sx+5. Their graphs intersect at (2,11), and p−q is a constant polynomial. Find r and s.
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p−q constant means slopes equal: s=3. Since p(2)=11:6+r=11⇒r=5. q(2)=6+5=11 also. Hence r=5,s=3.
Exam Strategy & Common Traps
| Topic | Best first step | Common trap |
|---|---|---|
| Degree | Identify the highest power after simplification. | Do not confuse coefficient size with degree. |
| Value of polynomial | Substitute the input carefully with brackets. | Sign errors with negative inputs. |
| Linear pattern | Check whether consecutive differences are constant. | Assuming every visible pattern is linear. |
| y=ax+b | Read a as slope and b as y-intercept. | Interchanging a and b. |
| Two-point model | Find constant change per unit first. | Using only output difference without dividing by input difference. |
| Parallel lines | Compare slopes. | Thinking equal intercepts imply parallel lines. |
| Growth/decay | Look at the sign of the constant change/slope. | Ignoring the domain in real-life models. |
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