91Ó°ÊÓ

A parabola is a U-shaped curve that can open either upwards or downwards. It is defined by a quadratic function, which takes the form \[ y = ax^2 + bx + c \].
The most common type of parabola is the simple one, given by the equation \[ y = x^2 \], which opens upwards.
This parabolic curve has unique properties, such as symmetry around its axis of symmetry and having a single vertex that represents its minimum or maximum point.

• The axis of symmetry is a vertical line that runs through the vertex, dividing the parabola into two mirror-image halves.
• The vertex is the point where the parabola either reaches its minimum or maximum value, depending on the direction the parabola opens.

In the given exercise, the parabola is defined by \[ y = x^2 \].
We aim to find a point on this curve that minimizes the area of the triangle formed with two other given points.

The vertex of a quadratic function is a crucial point that can represent either the maximum or minimum value of the function.
The general equation for a quadratic function is \[ y = ax^2 + bx + c \].
To find the vertex, we use the vertex formula:

\[ x = -\frac{b}{2a} \]

This formula gives us the x-coordinate of the vertex.
For example, in the quadratic equation \[ y = x^2 + 5x + 10 \],
we can identify \( a = 1 \ b = 5 \ c = 10 \).
Plugging these values into the formula, we get:

\[ x = -\frac{5}{2(1)} = -\frac{5}{2} = -2.5 \]

This tells us that the x-coordinate of the vertex is \(-2.5\).
To find the corresponding y-coordinate, substitute \(-2.5\) back into the quadratic equation:

\[ y = (-2.5)^2 + 5(-2.5) + 10 \]
\[ y = 6.25 - 12.5 + 10 \]
\[ y = 6.25 - 2.5 = 3.75 \]

In our exercise, the parabola equation is
\[ y = x^2 \].
Substitute \( -2.5 \) into this equation to get:

\[ y = (-2.5)^2 = 6.25 \]
Therefore, the point
\(C(-2.5, 6.25)\) represents the vertex and also the point that minimizes the area of the triangle.

Calculating the area of a triangle is an essential part of the exercise.
The formula to calculate the area of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:

\[ \text{Area} = \frac{1}{2} \big| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \big| \]

This formula helps us find the area by using the coordinates of the three vertices.
In this exercise, we have:

• Point A: (0, -10 )
• Point B: (2, 0 )
• Point C: ( x, x^2 )

Substitute these coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \big| 0(0 - x^2) + 2(x^2 + 10) + x(10 - 0) \big| \]

This simplifies to:

\[ \text{Area} = \frac{1}{2} \big| 2x^2 + 20 + 10x \big| \]
Given \( 2x^2 + 10x + 20 \) is always positive in the necessary range, we simplify it further to:

\[ \text{Area} = x^2 + 5x + 10 \]

Using the vertex formula, we find the point on the parabola where this area is minimized.
This final step gives us the specific coordinates of point C such that the area of triangle ABC is minimized.