A Cartesian framework

1. In a Cartesian framework, the functions f and g we know that:
the function (f) is defined by f (x) = 2x2, 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) = ax2, 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)=2x2, g(x)=x+3  f(x)=g(x) 2x2=x+3  2x2x3=0  a=2;b=1;c=3 D=b24ac=1242(3)=25 D>0  x1,2=2ab±D=41±25 x1,2=41±5 x1,2=0.25±1.25 x1=1.5 x2=1   Factored form of the equation:  2(x1.5)(x+1)=0  Ax=Ax=x1=1.5

Our quadratic equation calculator calculates it.

C=g(x)x=0 Cx=Cx=0
O=(0,0) a=dist(O,A)=(OxAx)2+(OyAy)2=(OxAx)2+(Oy29)2=(01.5)2+(04.5)24.7434 b=dist(O,C)=(OxCx)2+(OyCy)2=(00)2+(03)2=3 c=dist(A,C)=(AxCx)2+(AyCy)2=(AxCx)2+(29Cy)2=(1.50)2+(4.53)22.1213  s=(a+b+c)/2=(4.7434+3+2.1213)/24.9324 S=s (sa) (sb) (sc)=4.9324 (4.93244.7434) (4.93243) (4.93242.1213)=2.25

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