The table shows approximate values of x and y that satisfy a relationship of the form y = ab^x. Plot 10(y) against x to verify.

Question 3: The table shows approximate values of x and y that satisfy a relationship of the form y = ab^x. | x | 10 | 20 | 35 | 50 | 60 | 80 | |---|---|---|---|---|---|---| | y | 3.47 | 5.50 | 10.47 | 20.89 | 33.11 | 79.43 | Step 1: Linearize the relationship y = ab^x. Take the logarithm base 10 of both sides: _10(y) = _10(ab^x) Using logarithm properties, (MN) = M + N and (M^p) = p M: _10(y) = _10(a) + _10(b^x) _10(y) = _10(a) + x _10(b) This equation is in the form of a straight line Y = mX + C, where: Y = _10(y) X = x m = _10(b) (gradient) C = _10(a) (Y-intercept) Step 2: Create a new table with values for X=x and Y=_10(y). Calculate _10(y) for each given y value (rounded to 3 decimal places for plotting accuracy): | X = x | Y = _10(y) | |---|---| | 10 | _10(3.47) ≈ 0.540 | | 20 | _10(5.50) ≈ 0.740 | | 35 | _10(10.47) ≈ 1.020 | | 50 | _10(20.89) ≈ 1.320 | | 60 | _10(33.11) ≈ 1.520 | | 80 | _10(79.43) ≈ 1.900 | Step 3: Verify the relationship by drawing a linear graph. To verify the relationship y = ab^x, plot the points (X, Y) from the table above, i.e., plot _10(y) on the vertical axis against x on the horizontal axis. Choose appropriate scales for both axes to accommodate the range of values. Plot the six points: (10, 0.540), (20, 0.740), (35, 1.020), (50, 1.320), (60, 1.520), (80, 1.900). Draw a line of best fit through these plotted points. If the points lie approximately on a straight line, then the relationship y = ab^x is verified. Observing the calculated points, they appear to form a nearly straight line. Step 4: Find the approximate values of constants a and b using the graph. From the line of best fit drawn in Step 3, determine its gradient (m) and Y-intercept (C). Let's use two points from the line of best fit to calculate the gradient and Y-intercept. For demonstration, we'll use the first and last transformed data points as if they were read from the line of best fit: (X_1, Y_1) = (10, 0.540) and (X_2, Y_2) = (80, 1.900). Calculate the gradient m: m = (Y_2 - Y_1)/(X_2 - X_1) = (1.900 - 0.540)/(80 - 10) = (1.360)/(70) ≈ 0.019428 Calculate the Y-intercept C using the equation Y = mX + C and one of the points, e.g., (10, 0.540): 0.540 = (0.019428)(10) + C 0.540 = 0.19428 + C C = 0.540 - 0.19428 = 0.34572 Now, relate m and C back to a and b: m = _10(b) b = 10^m C = _10(a) a = 10^C Substitute the calculated values for m and C: b = 10^0.019428 ≈ 1.0456 a = 10^0.34572 ≈ 2.2166 Rounding the answers to 2 decimal places: a ≈ 2.22 b ≈ 1.05 The approximate values of the constants are a = 2.22, b = 1.05.