Introduction to Linear Polynomials | Exercise 2.6

Question 1

Draw the graphs of the following sets of lines. In each case, reflect on the role of $a$ and $b$ .

(i) $y = 4x, y = 2x, y = x$

(ii) $y = -6x, y = -3x, y = -x$

(iii) $y = 5x, y = -5x$

(iv) $y = 3x - 1, y = 3x, y = 3x + 1$

(v) $y = -2x - 3, y = -2x, y = 2x + 3$

Solution

Understand the Question

For any linear equation written in the slope-intercept form $y = ax + b$ :

Role of $a$ (Slope / Gradient): $a$ $a$ controls the steepness and direction of the line.
- If $a > 0$ , the line rises from left to right (positive slope). A larger value of $a$ means a steeper line.
- If $a < 0$ , the line falls from left to right (negative slope). A larger $|a|$ means a steeper downward slope.
- Lines with equal slope $a$ are parallel.
Role of $b$ ( $y$ -intercept): $b$ $b$ gives the point $(0, b)$ $(0, b)$ where the line intersects the $y$ $y$ -axis.
- It represents the vertical shift of the line.
- When $b = 0$ , the line passes directly through the origin $(0, 0)$ .

(i) Draw the graphs of $y = 4x, y = 2x, y = x$ . Reflect on the role of $a$ and $b$ .

Step 1 · Plot points and observe the role of positive $a$

Plot points for each line by finding coordinates $(x, y)$ :

For $y = 4x$ : If $x = 0$ , $y = 4(0) = 0 \implies (0, 0)$ If $x = 1$ , $y = 4(1) = 4 \implies (1, 4)$
For $y = 2x$ : If $x = 0$ , $y = 2(0) = 0 \implies (0, 0)$ If $x = 1$ , $y = 2(1) = 2 \implies (1, 2)$
For $y = x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = 1 \implies (1, 1)

Observations:

All three lines have $b = 0$ , so they all pass through the origin $(0, 0)$ .
As the positive value of $a$ increases ( $1 < 2 < 4$ ), the line becomes steeper.

Answer

(i) All lines pass through the origin $(0, 0)$ because $b = 0$ . A larger positive value of $a$ results in a steeper line.

(ii) Draw the graphs of $y = -6x, y = -3x, y = -x$ . Reflect on the role of $a$ and $b$ .

Step 1 · Plot points and observe the role of negative $a$

Plot points for each line:

For $y = -6x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = -6 \implies (1, -6)$
For $y = -3x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = -3 \implies (1, -3)$
For $y = -x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = -1 \implies (1, -1)

Observations:

All three lines have $b = 0$ , so they all pass through the origin $(0, 0)$ .
Since $a$ is negative, the lines slope downwards from left to right. A larger absolute value $|a|$ gives a steeper downward slope.

Answer

(ii) All lines pass through the origin $(0, 0)$ because $b = 0$ . Negative $a$ causes a downward slope, and a larger magnitude $|a|$ produces a steeper downward line.

(iii) Draw the graphs of $y = 5x, y = -5x$ . Reflect on the role of $a$ and $b$ .

Step 1 · Plot points and compare positive and negative $a$

Plot points for both lines:

For $y = 5x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = 5 \implies (1, 5)$
For $y = -5x$ : If $x = 0$ , $y = 0 \implies (0, 0)$ If $x = 1$ , $y = -5 \implies (1, -5)

Observations:

Both lines have $b = 0$ and pass through the origin $(0, 0)$ .
Both lines have the same magnitude $|a| = 5$ , giving them equal steepness, but they slope in opposite directions due to opposite signs.

Answer

(iii) Both lines pass through the origin $(0, 0)$ and have the same steepness since $|a| = 5$ , but they slope in opposite directions.

(iv) Draw the graphs of $y = 3x - 1, y = 3x, y = 3x + 1$ . Reflect on the role of $a$ and $b$ .

Step 1 · Plot points and observe parallel lines with varying $b$

Plot points for each line:

For $y = 3x - 1$ : If $x = 0$ , $y = 3(0) - 1 = -1 \implies (0, -1)$ If $x = 1$ , $y = 3(1) - 1 = 2 \implies (1, 2)$
For $y = 3x$ : If $x = 0$ , $y = 3(0) = 0 \implies (0, 0)$ If $x = 1$ , $y = 3(1) = 3 \implies (1, 3)$
For $y = 3x + 1$ : If $x = 0$ , $y = 3(0) + 1 = 1 \implies (0, 1)$ If $x = 1$ , $y = 3(1) + 1 = 4 \implies (1, 4)

Observations:

All three lines have identical slope $a = 3$ , meaning they are parallel to each other.
The constant $b$ shifts each line vertically, setting the $y$ -intercept at $(0, -1)$ , $(0, 0)$ , and $(0, 1)$ respectively.

Answer

(iv) All three lines are parallel because they have the same slope $a = 3$ . The constant $b$ vertically shifts the line and sets the $y$ -intercept.

(v) Draw the graphs of $y = -2x - 3, y = -2x, y = 2x + 3$ . Reflect on the role of $a$ and $b$ .

Step 1 · Plot points and compare lines with different slopes and intercepts

Plot points for each line:

For $y = -2x - 3$ : If $x = 0$ , $y = -2(0) - 3 = -3 \implies (0, -3)$ If $x = 1$ , $y = -2(1) - 3 = -5 \implies (1, -5)$
For $y = -2x$ : If $x = 0$ , $y = -2(0) = 0 \implies (0, 0)$ If $x = 1$ , $y = -2(1) = -2 \implies (1, -2)$
For $y = 2x + 3$ : If $x = 0$ , $y = 2(0) + 3 = 3 \implies (0, 3)$ If $x = 1$ , $y = 2(1) + 3 = 5 \implies (1, 5)

Observations:

Lines $y = -2x - 3$ and $y = -2x$ share the same slope $a = -2$ , so they are parallel.
Line $y = 2x + 3$ has slope $a = 2$ , so it slopes in the opposite direction.
The values of $b$ ( $-3, 0, 3$ ) determine where each line crosses the $y$ -axis.

Answer

(v) Lines $y = -2x - 3$ and $y = -2x$ are parallel with slope $-2$ . Line $y = 2x + 3$ slopes in the opposite direction with slope $2$ . The constant $b$ determines the $y$ -intercept.

Common Mistakes

Confusing Intercepts: Confusing the $y$ -intercept $(0, b)$ with the $x$ -intercept $(-b/a, 0)$ . Setting $x = 0$ gives the $y$ -intercept $b$ .
Sign of Slope: Misinterpreting a negative slope. A line with $a < 0$ slopes downwards from left to right, whereas $a > 0$ slopes upwards.
Parallel Condition: Assuming lines with equal magnitude slopes like $a = 5$ and $a = -5$ are parallel. Lines are only parallel if their slopes are strictly equal ( $a_1 = a_2$ ).