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, and 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, and 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)

Final Answer:

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

Step-by-step explanation:

$$\begin{gathered} f(x)=2x{}^{}2, \\ g(x)=x+3 \\ f(x)=g(x) \\ 2x^2=x+3 \\ 2x^2-x-3=0 \\ a=2;b=-1;c=-3 \\ D=b^2-4ac=1^2-4\cdot 2\cdot (-3)=25 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{1\pm \sqrt{25}}{4} \\ {x}_{1,2}=\frac{1\pm 5}{4} \\ {x}_{1,2}=0.25\pm 1.25 \\ {x}_1=1.5 \\ {x}_2=-1 \\ Ax=A_x=x_1=1.5 \end{gathered}$$

Our quadratic equation calculator calculates it.

$Ay=A_y=A_x+3=1.5+3=4.5$

$Bx=B_x=x_2=(-1)=-1$

$By=B_y=B_x+3=(-1)+3=2$

$$\begin{gathered} C=g(x)\cap x=0 \\ Cx=C_x=0 \end{gathered}$$

$Cy=C_y=C_x+3=0+3=3$

$$\begin{gathered} O=(0,0) \\ a=dist(O,A)=|OA|=\sqrt{(O_x-A_x{)}^2+(O_y-A_y{)}^2}=\sqrt{(0-1.5{)}^2+(0-4.5{)}^2}\doteq 4.7434 \\ b=dist(O,C)=|OC|=\sqrt{(O_x-C_x{)}^2+(O_y-C_y{)}^2}=\sqrt{(0-0{)}^2+(0-3{)}^2}=3 \\ c=dist(A,C)=|AC|=\sqrt{(A_x-C_x{)}^2+(A_y-C_y{)}^2}=\sqrt{(1.5-0{)}^2+(4.5-3{)}^2}\doteq 2.1213 \\ s=(a+b+c)/2=(4.7434+3+2.1213)/2\doteq 4.9324 \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{4.9324\cdot (4.9324-4.7434)\cdot (4.9324-3)\cdot (4.9324-2.1213)}=2.25 \end{gathered}$$

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