Now let us draw the graphs of y = −13x, y = −x, y = −3x by selecting suitable points on these lines.
Step-by-step solution
Idea: Pick two points on each line, choosing x-values that avoid fractions (for y = −⅓x, use multiples of 3). A negative coefficient makes y fall as x increases.
- Points:1 mark
Line Point 1 Point 2 Extra check y = −⅓x (0, 0) (3, −1) (−6, 2) y = −x (0, 0) (2, −2) (−3, 3) y = −3x (0, 0) (1, −3) (−1, 3) - Check one: for (3, −1) on y = −⅓x, −⅓ × 3 = −1 ✓.½ mark
- Plot each pair, join with a ruler and extend both ways (see the graph).1½ marks
- All three pass through (0, 0) and fall from left to right. y = −3x is the steepest, y = −x is equally inclined to both axes, and y = −⅓x is the least steep.1 mark
Check: Moving 1 unit right from the origin, the lines drop by ⅓, 1 and 3 units, matching their coefficients ✓.
Answer to write in the exam
y = −⅓x: x = 0 ⇒ y = 0; x = 3 ⇒ y = −1. Points (0, 0), (3, −1)
y = −x: points (0, 0), (2, −2)
y = −3x: points (0, 0), (1, −3)
Plot each pair of points and join with a straight line (graph)
∴ All three lines pass through (0, 0) and fall from left to right; y = −3x is the steepest
Common mistakes that cost marks
- Plotting (3, 1) instead of (3, −1) for y = −⅓x, losing the minus sign.
- Drawing the lines rising to the right. With a negative coefficient, y decreases as x increases.
- Thinking y = −⅓x is the steepest because −⅓ looks “bigger” than −3. Steepness depends on the size of the number: 3 > ⅓.
How this can come in the exam
The point (−2, 8) lies on the line
- y = 4x
- y = −4x
- y = −x
- y = −14x
Show answer
(B) y = −4x
−4 × (−2) = 8.
Draw the graphs of y = −2x and y = −12x on the same axes and state which is steeper.
Show answer
Points: (0, 0), (1, −2) and (0, 0), (2, −1) (1 mark). Correct graph (1 mark). y = −2x is steeper since 2 > ½; both pass through the origin and fall to the right (1 mark).Try one yourself
Give two points on y = −23x with whole-number coordinates.
Show answer
(0, 0) and (3, −2) (also (−3, 2)).
More questions like this
- The figure shows all the three graphs on the same axes. Does this help you to conclude anything about the linear equation y = − ax, a > 0, as a varies? What will happen when a > 1 and when a < 1?
- Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.
- Let us now draw the graphs of y = 2x − 1, y = 2x + 1, y = 2x + 5, first individually (as shown in the figure) and then on the same axes (as shown in the figure).
- Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
(Hint: In these equations a = 2, and b takes the values −1, 1 and 5, respectively.) - Now let us draw the graphs of the equations y = x + 3, y = 2x + 5 and y = 3x − 2. See the figure and observe where these lines cut the y-axis.