Chapter 2 · II · Algebra

Introduction to Linear Polynomials Worksheet

8 questions Polynomials and coefficients Linear growth and decay Slope, intercept and modelling

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Foundation

Expressions, polynomial definition, terms, coefficients and degree · 1 practice mark

Question 1

Which is a polynomial in x?

  1. 1/x + 2
  2. √x + 1
  3. 3x² − 2x + 5
  4. x⁻²
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(c) 3x² − 2x + 5

  1. A polynomial has non-negative integer exponents of its variable.
  2. 3x² − 2x + 5 has exponents 2, 1 and 0; the other options do not.

Slope, intercept and graphs including restricted domains · Linear patterns, growth, decay and modelling · 1 practice mark

Question 2

For y = −3x + 7, the slope and y-intercept are:

  1. 3 and 7
  2. −3 and 7
  3. 7 and −3
  4. −3 and −7
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(b) −3 and 7

  1. Compare y = mx + c with y = −3x + 7.
  2. Each increase of 1 in x lowers y by 3; x = 0 gives y = 7.

Expressions, polynomial definition, terms, coefficients and degree · 1 practice mark

Question 3

For P(x) = 4x³ − 2x + 9, the degree and coefficient of x² are:

  1. 3 and 0
  2. 2 and 4
  3. 3 and 2
  4. 9 and 0
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(a) 3 and 0

  1. The largest exponent with non-zero coefficient is 3.
  2. There is no x² term, so its coefficient is 0.

Zero of a linear polynomial · 1 practice mark

Question 4

The zero of P(x) = 7x − 4 is:

  1. −4/7
  2. 4/7
  3. 7/4
  4. 4
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(b) 4/7

  1. A zero makes the polynomial equal to zero.
  2. Solve 7x − 4 = 0 to obtain x = 4/7.

Practice

Expressions, polynomial definition, terms, coefficients and degree · Linear patterns, growth, decay and modelling · 3 practice marks

Question 5

A tank initially contains 120 litres and loses 5 litres per minute. Form a linear polynomial for the water remaining, state its degree and coefficients, and give the physically valid time interval.

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W(t) = 120 − 5t; degree 1; coefficients −5 and 120; 0 ≤ t ≤ 24 minutes.

  1. Subtract 5t from the initial 120 litres.
  2. The non-zero highest power of t is 1; the coefficient of t is −5 and the constant term is 120.
  3. Require t ≥ 0 and W(t) ≥ 0; solve 120 − 5t ≥ 0.

Slope, intercept and graphs including restricted domains · Linear patterns, growth, decay and modelling · 3 practice marks

Question 6

A taxi charges ₹40 plus ₹12 per kilometre. Form an equation and a table for distances 0, 1, 2 and 3 km. Describe the graph, slope and intercept.

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C = 12d + 40; costs ₹40, ₹52, ₹64, ₹76; slope 12 and intercept 40.

  1. The fixed charge is the constant term and the per-kilometre charge multiplies d.
  2. Substitute the four distances to form the table.
  3. Plot (0,40), (1,52), (2,64), (3,76). For continuously measured d ≥ 0 the graph is a ray on the line C = 12d + 40.

Linear cost model · 3 practice marks

Question 7

A phone plan costs ₹300 per month plus ₹8 per call-minute. Write C(m), find C(60), and interpret its slope and intercept.

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C(m) = 300 + 8m; C(60) = ₹780.

  1. Multiply the number of minutes m by the per-minute rate 8.
  2. Add the fixed charge 300; for m = 60, the total is 300 + 480 = 780.
  3. Slope 8 means ₹8 per additional minute; intercept 300 is the charge at zero minutes. The physical domain is m ≥ 0.

Challenge

Comparing linear models · 4 practice marks

Question 8

Two delivery charges are A(k) = 120 + 15k and B(k) = 200 + 10k rupees for k kilometres. Find the break-even distance and decide which is cheaper on each side of it.

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Break-even 16 km at ₹360; A is cheaper below 16 km, B above it.

  1. Set 120 + 15k = 200 + 10k.
  2. Subtract 10k and 120 to get 5k = 80, so k = 16.
  3. A(16) = B(16) = 360.
  4. A − B = 5k − 80. For 0 ≤ k < 16 it is negative; for k > 16 it is positive.