Example 15 - Chapter 2 Class 9 - Introduction to Linear Polynomials (Ganita Manjari

Transcript

Example 15 Draw the graphs of y = (−1)/3x, y = –x, y = –3x by selecting suitable points on these lines. To draw the graph, we join points which lie on the line For y = (−𝟏)/𝟑 x Putting x = 0 y = (−𝟏)/𝟑 × 0 y = 0 So, x = 0, y = 0 lie on the line i.e. (0, 0) lies on the line Putting x = 3 y = (−𝟏)/𝟑 × 3 y = –1 So, x = 3, y = –1 lie on the line i.e. (3, –1) lies on the line For y = –x Putting x = 0 y = –0 y = 0 So, x = 0, y = 0 lie on the line i.e. (0, 0) lies on the line Putting x = 1 y = –1 So, x = 1, y = –1 lie on the line i.e. (1, –1) lies on the line For y = –3x Putting x = 0 y = –3 × 0 y = 0 So, x = 0, y = 0 lie on the line i.e. (0, 0) lies on the line Putting x = 1 y = –3 × 1 y = –3 So, x = 1, y = –3 lie on the line i.e. (1, –3) lies on the line y = (−𝟏)/𝟑x y = –x y = –3x Thus, we observe that for line y = ax + b and a < 0 If a < –1 (like –2, –3, –10) the line becomes steeper than the standard y = –x line. If –1 < a < 0 (like (−1)/2 or (−1)/3), the line becomes less steep (flatter) than the y = –x line.