Geometry and Functions Flashcards

Equation of the Parabola (p):
- The parabola is defined by the quadratic equation in vertex form: $y = \frac{2}{3}(x-3)^2 - 3$ .
- The vertex $V$ of this parabola can be identified as $V(3 | -3)$ .
- The aperture factor (opening width) is $\frac{2}{3}$ , which means the parabola opens upwards and is slightly wider than the standard parabola (where $a=1$ ).
Equation of the Straight Line (g):
- The line is defined by the linear equation: $y = -0.5x + 0.5$ .
- The slope of the line is $m = -0.5$ , indicating it is a decreasing function.
- The y-intercept (the point where it crosses the y-axis) is at $c = 0.5$ , giving the point $(0 | 0.5)$ .
Basic Definitions and Domain:
- The set of points occurs in the Cartesian coordinate system defined by the total space $G = \mathbb{R} \times \mathbb{R}$ .
- The specific drawing task requires visualizing these functions over the interval $x \in [0; 8]$ .

Required Range for Visualization:
- The x-axis must cover at least the span from $0$ to $8$ .
- To accurately plot the parabola $p$ within this range, one should calculate key points:
  - For $x = 0$ : $y = \frac{2}{3}(0-3)^2 - 3 = \frac{2}{3}(9) - 3 = 6 - 3 = 3$ . Point: $(0 | 3)$ .
  - For $x = 3$ : $y = -3$ . Point: $(3 | -3)$ .
  - For $x = 6$ : $y = \frac{2}{3}(6-3)^2 - 3 = 6 - 3 = 3$ . Point: $(6 | 3)$ .
  - For $x = 8$ : $y = \frac{2}{3}(8-3)^2 - 3 = \frac{2}{3}(25) - 3 = \frac{50}{3} - 3 = 16.67 - 3 = 13.67$ . Point: $(8 | 13.67)$ .
- To accurately plot the line $g$ within this range:
  - For $x = 0$ : $y = 0.5$ . Point: $(0 | 0.5)$ .
  - For $x = 8$ : $y = -0.5(8) + 0.5 = -4 + 0.5 = -3.5$ . Point: $(8 | -3.5)$ .

Point Definitions:
- Points $A_n$ : These points lie on the parabola $p$ . Their coordinates are expressed as functions of the abscissa $x$ : $A_n(x | \frac{2}{3}(x-3)^2 - 3)$ .
- Points $C_n$ : These points lie on the line $g$ . Their coordinates share the same abscissa $x$ as $A_n$ : $C_n(x | -0.5x + 0.5)$ .
- Points $B_n$ : These points are defined by a specific geometric relationship to $A_n$ , forming triangles $A_n B_n C_n$ .
Movement Constraint for x:
- The triangles $A_n B_n C_n$ only exist for x-values in the open interval $x \in ]0.28; 7.05[$ . This range represents the values where the vertical distance between the line and the parabola allows for the triangle formation as described.
Vectorial Shift for Point B:
- The position of point $B_n$ is determined by the vector addition from point $A_n$ : $\vec{A_n B_n} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$
- This means to find $B_n$ , one must move $2$ units in the positive x-direction (right) and $5$ units in the positive y-direction (up) from point $A_n$ .
- Coordinates for $B_n$ : $B_n(x+2 | \frac{2}{3}(x-3)^2 - 3 + 5)$ , which simplifies to $B_n(x+2 | \frac{2}{3}(x-3)^2 + 2)$ .

Parameter for Construction:
- The task specifies the construction for the specific x-value: $x = 1.5$ .
Detailed Calculation of Vertices:
- Point $A_1$ :
  - $x = 1.5$
  - $y = \frac{2}{3}(1.5-3)^2 - 3 = \frac{2}{3}(-1.5)^2 - 3 = \frac{2}{3}(2.25) - 3 = 1.5 - 3 = -1.5$
  - Coordinates: $A_1(1.5 | -1.5)$
- Point $C_1$ :
  - $x = 1.5$
  - $y = -0.5(1.5) + 0.5 = -0.75 + 0.5 = -0.25$
  - Coordinates: $C_1(1.5 | -0.25)$
- Point $B_1$ (using vector shift from $A_1$ ):
  - $x = 1.5 + 2 = 3.5$
  - $y = -1.5 + 5 = 3.5$
  - Coordinates: $B_1(3.5 | 3.5)$
Drawing Instructions for $A_1 B_1 C_1$ :
- Locate $A_1(1.5 | -1.5)$ on the parabola.
- Locate $C_1(1.5 | -0.25)$ on the line, directly above $A_1$ on the same vertical line.
- Apply the vector $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ from $A_1$ to plot $B_1(3.5 | 3.5)$ .
- Connect the points $A_1$ , $B_1$ , and $C_1$ to form the triangle.