SPM Form 5 Add Math Project 2010 - Tugasan 1

[sekqy]

Notes: The Complete answer is available in the Premium Discussion Board. Register as a user now to see the full answer.(Registration is Free)

The purpose I create this post is for student to discuss their add math project. Copying is not encouraged. Just use the information here as a reference and then complete the work by yourself. Don't waste your time to email or personal message me to request for answer because I will not reply any such email.

If you have lost your question paper, you may download it from http://addmathsproject2010.blospot.com

Add Math Project 2010 Tugasan 1
Add Math Project 2010 Tugasan 2
Add Math Project 2010 Tugasan 3
Add Math Project 2010 Tugasan 4

I think they may provide answer as well later.

Discussion for other tugasan.
Tugasan 1, Tugasan 2, Tugasan 3, Tugasan 4

[sekqy]

Reminder: This is only a guide but not the complete answer. You should try to complete the project by yourself

a) Function I

tugasan 1-01.png (5.79 KiB) Viewed 36463 times

Maximum point (0,4.5) and pass through point (2,4)

y = a(x - b)^2  + c

b=0, c=4.5

y = a(x - 0)^2  + 4.5

y = ax^2  + 4.5 -  -  -  - (1)

Substitute (2,4) into (1)

4 = a(2)^2 + 4.5

4a + 4.5 = 4

4a = -0.5

a = -0.125

y = -0.125x^2 + 4.5

[sekqy]

Function (II)

tugasan 1-02.png (5.24 KiB) Viewed 36463 times

Maximum point (0,0.5) and pass through point (2,0)

y = a(x - b)^2  + c

b=0, c=0.5

y = a(x - 0)^2  + 0.5

y = ax^2  +0.5 -  -  -  - (2)

Substitute (2,0) into (2)

0 = a(2)^2 + 0.5

4a = -0.5

a = -0.125

y = -0.125x^2 + 0.5

[sekqy]

Function (III)

tugasan 1-03.png (5.72 KiB) Viewed 36463 times

Maximum point (2,4.5) and pass through point (0,4)

y = a(x - b)^2  + c

b=2, c=4.5

y = a(x - 2)^2  + 4.5  -  -  -  - (3)

Substitute (0,4) into (3)

4 = a(0-2)^2 + 4.5

4 = 4a + 4.5

4a = -0.5

a = -0.125

y = -0.125(x-2)^2 + 4.5

[sekqy]

(b)

tugasan 1-04.png (6.9 KiB) Viewed 36463 times

Area of the gray region
=Area of rectangle - Area P(Green Region)

= 4 \times 1 - 2\int_0^2 {( - 0.125x^2  + 0.5)dx}

= 4 - 2\left[ {\frac{{ - 0.125x^3 }}
{3} + 0.5x} \right]_0^2

= 4 - 2\left[ {\left( {\frac{2}
{3} - 0} \right)} \right]

=4 - \frac{4}
{3} = 2\frac{2}
{3}{\text{m}}^2

Notes:There was some typing error in the calculation above. The correct working should be

\begin{array}{l}
  = 4 \times 1 - 2\int_0^2 {( - 0.125x^2  + 0.5)dx}  \\ 
  = 4 - 2\left[ {\frac{{ - 0.125x^3 }}{3} + 0.5x} \right]_0^2  \\ 
  = 4 - 2(\frac{2}{3} - 0) = 4 - \frac{4}{3} = 2\frac{2}{3}m^2  \\ 
 \end{array}

The later part of this tugasan (Further Exploration (a), (b) and (c) is given in the Premium Discussion Board.

[skydream]

(b)

tugasan 1-04.png

Area of the gray region
=Area of rectangle - Area P(Green Region)

= 4 \times 1 - 2\int_0^2 {( - 0.125x^2  + 0.5)dx}

= 4 - 2\left[ {\frac{{ - 0.125x^3 }}
{3} + 0.5x} \right]_0^2

= 4 - 2\left[ {\left( {\frac{2}
{3} - 0} \right)} \right]

=4 - \frac{4}
{3} = 2\frac{2}
{3}{\text{m}}^2

picture like this how can i made in computer? can u teach me

[sekqy]

picture like this how can i made in computer? can u teach me

You can draw it by using powerpoint. Draw the rectangle, and then draw the circle, then draw another rectangle that has same colour as the background to cover the lower part of the circle, and then draw the graph. Done!

To export the picture, select all the objects, right click on it, select save as picture.

[skydream]

picture like this how can i made in computer? can u teach me

You can draw it by using powerpoint. Draw the rectangle, and then draw the circle, then draw another rectangle that has same colour as the background to cover the lower part of the circle, and then draw the graph. Done!

To export the picture, select all the objects, right click on it, select save as picture.

can u pm me ur msn?i have something about folio need to ask u

[sekqy]

can u pm me ur msn?i have something about folio need to ask u

You can post your question here.

[skydream]

can u pm me ur msn?i have something about folio need to ask u

You can post your question here.

graph of function u also use power point to do?