Can You Draw the Graphic Representation of a Function and Geometry Problem?

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Homework Statement



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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


Homework Equations





The Attempt at a Solution

 
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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)

On the X axis, we consider points A, B and H of respective abscissas 3, -3 and x (x belongs R)
Did you repost this? I remember responding to post where both A and B had abscissa 3! It would have been better to add your correction to the same thread.

Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x
Okay so C also has abscissa (x-value) 3 and is on the graph y= (2/3)x+ 2 so y= (2/3)(3)+ 2= 4. C is the point (3, 4). Similarly, M is the point (x, (2/3)x+ 2).

1°) Make the complete figure ?
Surely, if you are doing a problem like this, you know that the graph of a function like this is a straight line? When x= -3 (as with point B), y= (2/3)(-3)+ 2= -2+ 2= 0. The graph is the straight line that passes through (-3, 0) and (3, 4).

2°) Calculate the distances AH and MH according to x and by means of the absolute values ?
Since A is (3, 0) and H is (x, 0), the distance from A to H is |x-(-3)|= |x+ 3|. Since M is the point (x, (2/3)x+ 2 and, again, H is (x, 0), the distance from M to H is |x- (2/3)x0- 2|= |(1/3)x- 2|.

3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?
CHA is a triangle with vertices at (2, 4), (x, (2/3)x+ 2), and (2, 0). That's a right triangle with base of length AH and height HC= |4- 2|= 2. The area of a triangle is (1/2)*base*height. The area is a formula involving x.

MOB (O is the origin, (0,0)?) is not a right triangle but it is still a triangle with base OB= 3 and and height AC. Again, you will get a formula involving x. If the areas are equal, set those two formulas equal and solve for x.

Please Anyone for help me, Can u drwan this graphic please please and represents all points A B H M and C
As I said above, the graph is a straight line passing through (-3, 0) and (3, 4).


Homework Equations





The Attempt at a Solution

 
Function Geometry

Hello Thank you for your answer but can you make this graphic please He is going to help me throughout my homework By representing all the points PLEASE PLEASE PLEASE
 
Function ax+b

Hello Thank you for your answer but can you make this graphic please He is going to help me throughout my homework By representing all the points PLEASE PLEASE PLEASE
 
No, I am not going to draw a straight line for you! I have already told you that the graph is the straight line passing (-3, 0) and (3, 4). If you can't draw that you need more help than I can give you.
 
pLEASE BUT WHEN I DREW THIS CURVE I DONT OBTAIN TWO TRIANGLES PLEASE HELP ME TO DRAW THE CURVE
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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