A Cartesian framework

1. In a Cartesian framework, the functions f and g we know that:
the function (f) is defined by f (x) = 2x ^ 2, the function (g) is defined by g (x) = x + 3, the point (O) is the origin of the reference, point (C) is the point of intersection of the graph of the function (g) with the ordinate axis, points A and B are the points of intersection of the graphs of the functions (f) and (g)
1.1 write the coordinates of points (A) and (B)
1.2. indicate the solutions of the equation f (x) = g (x)
1.3. determine the area of the triangle [OAC]
2.1. In the figure, part of the graph of a quadratic function (f) of the type: f (x) = ax ^ 2, a related function (g) and the trapezium [OBAC] are represented in a Cartesian framework. :
point (O) is the origin of the reference, point (B) is the point of intersection of the graph of the function (g) with the ordinate axis equal to 6, point A is the point of the intersection of the graphs of the functions (f) and (g)
point (C) belongs to the abscissa axis and has abscissa equal to 4, the trapezium area [OBAC] is equal to 18
2.1 Determine the coordinates of the point (A)
2.2. Determine the algebraic expressions of the functions (f) (g)

Correct answer:

Ax = 1.5
Ay = 4.5
Bx = -1
By = 2
Cx = 0
Cy = 3
S = 2.25

Step-by-step explanation:

f (x) = 2 x^{2}, g (x) = x + 3 f (x) = g (x) 2 x^{2} = x + 3 2 x^{2} - x - 3 = 0 a = 2; b = - 1; c = - 3 D = b^{2} - 4 a c = 1^{2} - 4 \cdot 2 \cdot (- 3) = 25 D > 0 x_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a} = \frac{1 \pm \sqrt{25}}{4} x_{1, 2} = \frac{1 \pm 5}{4} x_{1, 2} = 0.25 \pm 1.25 x_{1} = 1.5 x_{2} = - 1 Factored form of the equation: 2 (x - 1.5) (x + 1) = 0 A x = A_{x} = x_{1} = 1.5 = \frac{3}{2} = 1 \frac{1}{2}

Our quadratic equation calculator calculates it.

A y = A_{y} = A_{x} + 3 = 1.5 + 3 = \frac{9}{2} = 4 \frac{1}{2} = 4.5

B x = B_{x} = x_{2} = (- 1) = - 1

B y = B_{y} = B_{x} + 3 = (- 1) + 3 = 2

C = g (x) \cap x = 0 C x = C_{x} = 0

C y = C_{y} = C_{x} + 3 = 0 + 3 = 3

O = (0, 0) a = d i s t (O, A) = \sqrt{(O_{x} - A_{x})^{2} + (O_{y} - A_{y})^{2}} = \sqrt{(0 - 1.5)^{2} + (0 - 4.5)^{2}} ≐ 4.7434 b = d i s t (O, C) = \sqrt{(O_{x} - C_{x})^{2} + (O_{y} - C_{y})^{2}} = \sqrt{(0 - 0)^{2} + (0 - 3)^{2}} = 3 c = d i s t (A, C) = \sqrt{(A_{x} - C_{x})^{2} + (A_{y} - C_{y})^{2}} = \sqrt{(1.5 - 0)^{2} + (4.5 - 3)^{2}} ≐ 2.1213 s = (a + b + c) / 2 = (4.7434 + 3 + 2.1213) / 2 ≐ 4.9324 S = \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c)} = \sqrt{4.9324 \cdot (4.9324 - 4.7434) \cdot (4.9324 - 3) \cdot (4.9324 - 2.1213)} = \frac{9}{4} = 2 \frac{1}{4} = 2.25

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