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Hi, for the 2^x + k graph, you can substitute any point on the curve to find out what the value of k is. For simplicity, I substitute the point (0,0) into y = 2^x + k since (0,0) is a point on this curve. Then I will get 0 = 1 + k which gives me the value of k, which is -1.
For the straight line graph, y = mx + c, simply fall back on your Sec 1 Math knowledge - calculate the gradient of the straight line by using any two points shown in the graph, and you will get m = 1 (since m here represents the gradient). Similarly, you can derive c graphically by seeing where the y intercept of the line is. The line cuts the y-axis at 2, so c = 2 (since c here represents the y-intercept).
As for the second part, on showing that the points of intersection will satisfy 2^x - x = 3, simply equate the two equations together: 2^x - 1 = x + 2. Then rearrange this equation to get 2^x - x = 3.
For the straight line graph, y = mx + c, simply fall back on your Sec 1 Math knowledge - calculate the gradient of the straight line by using any two points shown in the graph, and you will get m = 1 (since m here represents the gradient). Similarly, you can derive c graphically by seeing where the y intercept of the line is. The line cuts the y-axis at 2, so c = 2 (since c here represents the y-intercept).
As for the second part, on showing that the points of intersection will satisfy 2^x - x = 3, simply equate the two equations together: 2^x - 1 = x + 2. Then rearrange this equation to get 2^x - x = 3.