How Does f(x)=2/3x+2 Define Geometry on the Cartesian Plane?

shawqidu19
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Homework Statement



The function defined on R by f(x)= 2/3x+2

And cf his graphic representation in an orthonormal mark (o; i; j)

On the X axis, we consider points A, B and H of respective abscissas 3, 3 and x (x belongs R)

Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x


1°) Make the complete figure ?

2°) Calculate the distances AH and MH according to x and by means of the absolute values ?

3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?

Homework Equations





The Attempt at a Solution



You can make me the graphic representation and the rest I shall make him(it) alone please make for me please
 
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For clarification, by f(x)= 2/3x+2, do you mean 2/(3x)+ 2 or 2/(3x+2) or (2/3)x+ 2. I suspect the last.

I am not at all clear what you mean by "cf his graphic representation in an orthonormal mark (o; i; j)". o is the origin, i the x-axis, j the y-axis?
 
HallsofIvy said:
For clarification, by f(x)= 2/3x+2, do you mean 2/(3x)+ 2 or 2/(3x+2) or (2/3)x+ 2. I suspect the last.

I am not at all clear what you mean by "cf his graphic representation in an orthonormal mark (o; i; j)". o is the origin, i the x-axis, j the y-axis?

I mean (2/3)x+2 and o is the origin i the x-axis j the y-axis

Cf is the name of the curve which represents f(x) = (2/3)x+2

Can you make this representation please please...

I ask to you : Can you make this graphic representation ? please just the graphic
 
Last edited:
shawqidu19 said:

Homework Statement



The function defined on R by f(x)= 2/3x+2

And cf his graphic representation in an orthonormal mark (o; i; j)
Assuming this is f(x)= (2/3)x+ 2, then its graph is a straight line, passing through (0, 2) and (3, 4).

On the X axis, we consider points A, B and H of respective abscissas 3, 3 and x (x belongs R)
?? A and B are on the x-axis and both have abscissa 3? Then A and B are the same point?

Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x
I'm not certain I am interpreting this right but I think, since A= (3, 0), C= (3, 4).
Since H= (x,0), M= (x, (2/3)x+ 2).

1°) Make the complete figure ?
As I said above, draw the straight line passing through points (0, 2) and (2, 4).

2°) Calculate the distances AH and MH according to x and by means of the absolute values ?
Okay. A= (2, 0) and H= (x, 0) so AH= |x-2|. M= (x, (2/3)x+ 2). MH= |x- (2/3)x-2|= |(1/3)x- 2|

3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?
The triangle CHA is a right triangle with height AH and Base AC- and AC= |4-2|= 2.
Area is (1/2)base*height

The triangle MOB (O is the origin, (0, 0)?) is not a right triangle with but we can still calculate its height MH above, and base OH= x. Again area= (1/2)base*height.

Set the two areas equal and solve for x.

Homework Equations





The Attempt at a Solution



You can make me the graphic representation and the rest I shall make him(it) alone please make for me please[/QUOTE]
 
Function

A has got 3 as abscissa and B has got -3 as abscissa i am sorry i mistake
 
Fucntion geometry

Please can you draw me this curve please please and alla other questions i will do alone thank you help me.
 
Function and Geometry (Part 2 )

The function defined on R by f(x)= 2/3x+2

And cf his graphic representation in an orthonormal mark (o; i; j)

On the X axis, we consider points A, B and H of respective abscissas 3, -3 and x (x belongs R)

Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x


1°) Make the complete figure ?

2°) Calculate the distances AH and MH according to x and by means of the absolute values ?

3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?


Please Anyone for help me, Can u drwan this graphic please please and represents all points A B H M and C
 
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