Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.

Step-by-step solution

Idea: In y = ax + b, a is the slope (sign = direction, size = steepness) and b is where the line cuts the y-axis.

1. Points: y = 3x + 1 passes through (0, 1) and (1, 4); y = −3x + 1 passes through (0, 1) and (1, −2) (see the graph).1 mark
2. Same: both have b = 1, so both cut the y-axis at (0, 1) (y-intercept 1). Both have slope of size 3, so they are equally steep.1 mark
3. Different: y = 3x + 1 has slope +3 and rises from left to right (linear growth); y = −3x + 1 has slope −3 and falls (linear decay). Each is the mirror image of the other in the y-axis, and they meet only at (0, 1).1 mark

Check: At x = 2: 3(2) + 1 = 7 and −3(−2) + 1 = 7, so the point (2, 7) on one line matches (−2, 7) on the other, confirming the mirror image ✓.

Answer to write in the exam

Common mistakes that cost marks

• Saying the lines are parallel. Parallel lines have equal slopes; here the slopes are 3 and −3.
• Saying they cut the y-axis at different points. Both have b = 1.
• Saying y = −3x + 1 is less steep. The size of the slope is 3 for both.

How this can come in the exam

Try one yourself

How do the graphs of y = 12x + 2 and y = −12x + 2 differ?

More questions like this

• 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.
• Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
• Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.

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