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Quadratic functions are polynomial functions of degree 2, characterized by the general form: y = ax^2 + bx + c. In the given exercise, both distance functions, d1(t) and d2(t), are quadratic functions. For the faster car, the equation is d1(t) = 10t^2. This function describes how the distance traveled by the faster car changes over time (t). As t increases, the distance covered grows quadratically. For the slower car, the distance function is d2(t) = 5(t + 2)^2. To better understand, expand it to also get a quadratic form: 5(t + 2)^2 = 5(t^2 + 4t + 4) = 5t^2 + 20t + 20. This tells us how the slower car's distance changes over time, but also highlights the handicap since it starts with an additional +2 seconds in the t variable. Quadratic functions are essential in modeling scenarios where acceleration is constant, like the drag racing example.

Distance-time relationships describe how physical objects move over time. In the provided problem, we see two distance-time relationships. The faster car's movement is described by d1(t) = 10t^2, showing that its distance increases as a function of time. The coefficient 10 here implies that the car is accelerating rapidly. The slower car’s movement is given by d2(t) = 5(t + 2)^2, a bit more complex because of the handicap time (t+2). When expanded, it becomes d2(t) = 5t^2 + 20t + 20, which helps us to understand its motion better by seeing its instantaneous speed and acceleration more clearly. Understanding distance-time relationships is crucial for calculating and predicting movements of objects, whether for sports statistics, travel planning, or physics problems.

The concept of the difference of functions is vital in situations where we need to understand how one variable changes relative to another. Here, we need the relative distance between two cars. We write the relative distance function, h(t), by subtracting the faster car’s distance function, d1(t), from the slower car’s distance function, d2(t). h(t) = d2(t) - d1(t) By substituting the given functions, d1(t) = 10t^2 and d2(t) = 5(t + 2)^2, we get: h(t) = 5(t + 2)^2 - 10t^2. This function h(t) shows us the distance gap between the two cars at any given time. Understanding the difference of functions allows one to analyze and compare varying quantities efficiently.

Expansion and simplification in algebra involve rewriting expressions to make them easier to work with. In the context of this problem, we begin with: h(t) = 5(t + 2)^2 - 10t^2. We expand the term 5(t + 2)^2 to simplify the relative distance function: 5(t + 2)^2 = 5(t^2 + 4t + 4) = 5t^2 + 20t + 20. This gives us: h(t) = 5t^2 + 20t + 20 - 10t^2. Next, we combine like terms by subtracting 10t^2 from 5t^2: h(t) = -5t^2 + 20t + 20. Expansion and simplification make complex expressions clearer and are foundational skills in algebra, enabling more comfortable manipulation and understanding of mathematical functions.