91Ó°ÊÓ

Defining data points is an essential first step in solving any problem involving a quadratic function. In this exercise, we are provided with pairs of travel speeds (in km/h) and the number of daytime accidents (for every 200 million km driven). For example:

• (60, 100)
• (80, 130)
• (100, 200)

Each pair consists of an x-value (travel speed) and a y-value (number of accidents). Identifying these points correctly ensures accurate formulation of equations later on.

Solving a system of equations is crucial for finding the coefficients of the quadratic function. In our example, we form three equations from the data points:
1. For (60, 100): \[ 100 = 3600a + 60b + c \]
2. For (80, 130): \[ 130 = 6400a + 80b + c \]
3. For (100, 200): \[ 200 = 10000a + 100b + c \]
We then simplify these to:

• 3600a + 60b + c = 100
• 6400a + 80b + c = 130
• 10000a + 100b + c = 200

Next, we solve this system using either elimination or substitution methods.

Function estimation involves using our derived quadratic function to predict unknown values. After solving our system of equations, we get the function:
\[ f(x) = 0.05x^2 - 5.5x + 250 \]
To estimate the number of daytime accidents at a particular speed, we simply insert the speed value into our function. For 50 km/h:
\[ f(50) = 0.05(50)^2 - 5.5(50) + 250 \]
Resulting in:
\[ f(50) = 125 - 275 + 250 = 100 \]
Therefore, at 50 km/h, approximately 100 accidents occur for every 200 million km driven.

Substitution and elimination methods are techniques used to solve systems of linear equations. The elimination method involves adding or subtracting equations to eliminate one of the variables. For example, from our simplified equations:
\[ 6400a + 80b + c - 3600a - 60b - c = 30 \]
This simplifies to:
\[ 2800a + 20b = 30\]
We use it again to find another equation:
\[ 10000a + 100b + c - 6400a - 80b - c = 70 \]
This simplifies to:
\[ 3600a + 20b = 70\]
The substitution method involves solving one equation for a variable and substituting that value into another equation. For instance, solving:
\[ 800a = 40 \]
Gives us:
\[ a = 0.05 \]
Then we substitute this value into another simplified equation to find:
\[ b = -5.5, \]
And finally, we find the last coefficient c. Both methods are valuable tools in solving systems of equations.