Step 1
Given function \(N = 61.86t^2 − 237.43t + 943.51\)
1) For finding the vertical intercept, put \(t = 0\)
\(N = 61.86(0)^2 − 237.43(0) + 943.51\)

\(N = 943.51\) The practical meaning of the vertical intercept in this situation is: When\(t = 0,\) that is in 1990, the number of scooters sold in India = 943 (approximately). 2) When N = 1000, the equation becomes, \(61.86t^2 − 237.43t + 943.51 = 1000\) Solving using bisection method, \(f(t) = 61.86t^2 − 237.43t + 943.51 − 1000\) \(= 61.86t^2 − 237.43t − 56.49\) Step 2 \(f(−1) = 61.86+237.43−56.49 = 242.8\)

\(f(0) = −56.49\)

\(f(1) = 61.86 − 237.43 − 56.49 =−232.06\) Since \(f(−1)>0 and f(0)<0\)</span>, zero of the function lies between -1 and 0. Take \(t1=\frac{−1 + 0}{2}} =\frac{−1}{2} = −0.5\) \(f(−0.5) = 15.465 + 118.715 − 56.49\)

\(=77.69 > 0\) Thus root lies between -0.5 and 0 Take \(t_2 = \frac{−0.5 + 0}{2} = −0.25\)

\(f(−0.25) = 3.86 + 59.36 − 56.49\)

\(=6.73 > 0\) Step 3 Thus root lies between -0.25 and 0 Take \(t_3 = \frac{−0.25 + 0}{2} = −0.125\)

\(f(−0.125) = 0.96+29.69−56.49\)

\(=−25.84<0\)</span> \(\begin{array}{|c|c|}n & a (negative) & b (positive) & t=a+b2 & f(t)\ 1 & -1 & 0 & -0.5 & 77.69\\ 2 & -0.5 & 0 & -0.25 & 6.73\\ 3 & -0.25 & 0 & -0.125 & -25.84\end{array}\) Root here lies between -0.25 and -0.125. Thus root lies between \(\frac{−0.25−0.125}{2} = −0.1875\).

\(N = 943.51\) The practical meaning of the vertical intercept in this situation is: When\(t = 0,\) that is in 1990, the number of scooters sold in India = 943 (approximately). 2) When N = 1000, the equation becomes, \(61.86t^2 − 237.43t + 943.51 = 1000\) Solving using bisection method, \(f(t) = 61.86t^2 − 237.43t + 943.51 − 1000\) \(= 61.86t^2 − 237.43t − 56.49\) Step 2 \(f(−1) = 61.86+237.43−56.49 = 242.8\)

\(f(0) = −56.49\)

\(f(1) = 61.86 − 237.43 − 56.49 =−232.06\) Since \(f(−1)>0 and f(0)<0\)</span>, zero of the function lies between -1 and 0. Take \(t1=\frac{−1 + 0}{2}} =\frac{−1}{2} = −0.5\) \(f(−0.5) = 15.465 + 118.715 − 56.49\)

\(=77.69 > 0\) Thus root lies between -0.5 and 0 Take \(t_2 = \frac{−0.5 + 0}{2} = −0.25\)

\(f(−0.25) = 3.86 + 59.36 − 56.49\)

\(=6.73 > 0\) Step 3 Thus root lies between -0.25 and 0 Take \(t_3 = \frac{−0.25 + 0}{2} = −0.125\)

\(f(−0.125) = 0.96+29.69−56.49\)

\(=−25.84<0\)</span> \(\begin{array}{|c|c|}n & a (negative) & b (positive) & t=a+b2 & f(t)\ 1 & -1 & 0 & -0.5 & 77.69\\ 2 & -0.5 & 0 & -0.25 & 6.73\\ 3 & -0.25 & 0 & -0.125 & -25.84\end{array}\) Root here lies between -0.25 and -0.125. Thus root lies between \(\frac{−0.25−0.125}{2} = −0.1875\).