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:

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

Our quadratic equation calculator calculates it.

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

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

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

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

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

$$\begin{gathered} O=(0,0) \\ a=dist(O,A)=\sqrt{(O_x-A_x{)}^2+(O_y-A_y{)}^2}=\sqrt{(O_x-A_x{)}^2+(O_y-\frac{9}{2}{)}^2}=\sqrt{(0-1.5{)}^2+(0-4.5{)}^2}\doteq 4.7434 \\ b=dist(O,C)=\sqrt{(O_x-C_x{)}^2+(O_y-C_y{)}^2}=\sqrt{(0-0{)}^2+(0-3{)}^2}=3 \\ c=dist(A,C)=\sqrt{(A_x-C_x{)}^2+(A_y-C_y{)}^2}=\sqrt{(A_x-C_x{)}^2+(\frac{9}{2}-C_y{)}^2}=\sqrt{(1.5-0{)}^2+(4.5-3{)}^2}\doteq 2.1213 \\ s=(a+b+c)/2=(4.7434+3+2.1213)/2\doteq 4.9324 \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{4.9324\cdot (4.9324-4.7434)\cdot (4.9324-3)\cdot (4.9324-2.1213)}=2.25 \end{gathered}$$

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