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A quadratic function typically has a parabolic shape that will either open upwards or downwards. When the parabola opens upwards, like a smile, it has a lowest point called the minimum value. For the given function, since both constant coefficients \(a_1\) and \(a_2\) are positive, each part of the function is a parabola opening upwards. This means the entire function also opens upward and has a minimum value.

The task involves finding the exact \(x\) position where this minimum occurs. Knowing where these minima lie allows us to determine where the function reaches its lowest point, helping in understanding the behavior of quadratic functions in terms of optimization. In this case, by solving derivative equalities, we found the precise \(x\) where the function achieves its minimum.

Calculating the derivative is essential when we need to determine the behavior of functions, such as finding minimum or maximum values. A derivative gives us a formula that describes the slope of the function at any point. By finding this slope, we're able to see where the function levels out and doesn't rise or fall. These points are typically minimum or maximum values.

For our specific function, we calculated the derivative with respect to \(x\) and ended up with:

• \(\frac{dy}{dx} = 2a_1(x-x_1) + 2a_2(x-x_2)\)

Each term in this expression comes from differentiating the squared terms \((x-x_1)^2\) and \((x-x_2)^2\), using the chain rule. This derivative tells us how the function \(y\) changes as \(x\) changes.

After finding the derivative, the next step is identifying critical points by setting the derivative equal to zero. Critical points occur where the slope of the function is zero, indicating that there could be a minimum or maximum value at that point.

Solving the equation:

• \(2a_1(x-x_1) + 2a_2(x-x_2) = 0\)

gives us the value of \(x\) where the slope is zero. After simplification, it leads to the conclusion:

• \( (a_1 + a_2)x = a_1x_1 + a_2x_2 \)

This critical point calculation shows us where the function could have a promising minimum point, especially since we know those leading coefficients \(a_1\) and \(a_2\) are positive.

The function in question is composed of quadratic expressions, which are essentially parabolas characterized by terms of the form \((x-x_i)^2\). When written generally, a quadratic expression is given by:

• \(ax^2 + bx + c\)

where \(a\), \(b\), and \(c\) are constants. Here, \(a_i(x-x_i)^2\) serves as a focus through defining how deeply or shallowly the parabola curves and at what \(x_i\) it reaches its minimum.

In our exercise, the function combines two such parabolas, each anchoring its minimum at points \(x_1\) and \(x_2\). The combined expression weighs these parabolic curves based on \(a_1\) and \(a_2\), dictating the mutual influence of each parabolic form. Understanding these components is crucial in analyzing the parabolic graphs for minimum values and possible optimisation.