The value of Bob’s car can be calculated from the formula $$V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500$$ where $V$ is the value of the car in pounds (£) and $t$ is the age in years - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 5

To find $t$ when $V = 9500$, set up the equation:

$9500 = 17000e^{-0.25t} + 2000e^{-0.5t} + 500.$

First, simplify the equation:

ightarrow 9000 = 17000e^{-0.25t} + 2000e^{-0.5t}.$$ Rearranging gives: $$17000e^{-0.25t} + 2000e^{-0.5t} = 9000.$$ To solve, we can use the substitution $x = e^{-0.25t}$, leading to the equation: $$17000x + 2000x^2 = 9000.$$ This simplifies to a quadratic equation: $$2000x^2 + 17000x - 9000 = 0.$$ Using the quadratic formula, we can find the roots and solve for $t$.

To find the rate of decrease at $t = 8$, we need to differentiate the value function:

$V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500.$ Using the chain rule, the derivative is:

$\frac{dV}{dt} = -4250e^{-0.25t} - 1000e^{-0.5t}.$ Substituting $t = 8$ into the derivative gives:

$\frac{dV}{dt} = -4250e^{-0.25(8)} - 1000e^{-0.5(8)}.$ Calculating this will give us the answer, which we will approximate to the nearest pound.