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Linear relationships · 4 marks

The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

  1. (i) Find the polynomial p(x).
  2. (ii) Find the coordinates where the graph of p(x) cuts the axes.
  3. (iii) Draw the graph of p(x) and verify your answers.
Answer: (i) p(x) = 3x + 2 (ii) y-axis at (0, 2), x-axis at (−23, 0) (iii) the line through (1, 5) and (3, 11) passes through both these points.

Step-by-step solution

Idea: Write p(x) = ax + b. Each point gives an equation: p(1) = 5 and p(3) = 11. Solve for a and b.

xy−3−2−1123−4−2246810120y = 3x + 2(1, 5)(3, 11)(0, 2)(−⅔, 0)

(i) Find the polynomial p(x).

  1. Let p(x) = ax + b. Then a + b = 5 … (1) and 3a + b = 11 … (2).½ mark
  2. From (1), b = 5 − a. In (2): 3a + 5 − a = 11 ⇒ 2a = 6 ⇒ a = 3, and b = 2. So p(x) = 3x + 2.1 mark
p(x) = 3x + 2

(ii) Find the coordinates where the graph of p(x) cuts the axes.

  1. y-axis (x = 0): p(0) = 2, so (0, 2).½ mark
  2. x-axis (y = 0): 3x + 2 = 0 ⇒ x = −23, so (−23, 0).½ mark
(0, 2) and (−23, 0)

(iii) Draw the graph of p(x) and verify your answers.

  1. Plot (1, 5) and (3, 11), join them and extend the line (see the graph).1 mark
  2. The line crosses the y-axis at 2 and the x-axis between −1 and 0, at about −0.67 = −23, which agrees with (ii).½ mark
The graph confirms (0, 2) and (−23, 0).
(i) p(x) = 3x + 2 (ii) (0, 2) on the y-axis and (−23, 0) on the x-axis (iii) the graph through (1, 5) and (3, 11) confirms both points.

Check: p(1) = 3 + 2 = 5 ✓, p(3) = 9 + 2 = 11 ✓, p(−23) = −2 + 2 = 0 ✓.

Answer to write in the exam

(i)

Let p(x) = ax + b

a + b = 5 … (1); 3a + b = 11 … (2)

(2) − (1): 2a = 6 ⇒ a = 3; b = 5 − 3 = 2

∴ p(x) = 3x + 2

(ii)

x = 0: y = 2 ⇒ (0, 2)

y = 0: 3x + 2 = 0 ⇒ x = −23 ⇒ (−23, 0)

∴ Cuts the y-axis at (0, 2) and the x-axis at (−23, 0)

(iii)

Points (1, 5), (3, 11), (0, 2) plotted and joined (graph)

∴ From the graph, the line cuts the axes at (0, 2) and (−23, 0), as found

Common mistakes that cost marks

  • Taking the slope as 3 − 111 − 5 = 13 (the change in x over the change in y). The slope is change in y ÷ change in x = 62 = 3.
  • Giving the x-axis point as (0, −23). On the x-axis y = 0, so the point is (−23, 0).
  • Mixing up where to put 0: on the y-axis, x = 0; on the x-axis, y = 0.

How this can come in the exam

MCQ (1 mark)

The linear polynomial whose graph passes through (0, −1) and (2, 7) is

  1. 4x − 1
  2. 3x + 1
  3. 4x + 1
  4. −x + 4
Show answer

(A) 4x − 1
b = −1 and 2a − 1 = 7 ⇒ a = 4.

Short answer (3 marks)

The graph of a linear polynomial passes through (2, 1) and (5, 10). Find the polynomial and where its graph cuts the x-axis.

Show answer2a + b = 1, 5a + b = 10 ⇒ 3a = 9, a = 3 (1 mark); b = −5: p(x) = 3x − 5 (1 mark). x-axis: x = 53, point (53, 0) (1 mark).

Try one yourself

Find the linear polynomial whose graph passes through (−1, 4) and (2, −2), and its intercept on the y-axis.

Show answer

3a = −6 ⇒ a = −2, b = 2: p(x) = −2x + 2; it cuts the y-axis at (0, 2).

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