The Hardest Goodbye

I had always dreaded goodbyes, feared separating from my family all my life and was destined to face the hardest separation of all. Since the day I started going to boarding school, my only fear in life was separation as I went to boarding school at the age of six. At my first boarding school, I had my older sisters with me and two years later my younger brother joined the boys’ school and I was able to see him during school days as the young boys came daily to our school until grade 3. I remember that I spent all my breaks with him and felt very sad that he had to leave at the end of the day while I and my two other sisters were fortunate to stay together. It broke my heart to see him leave every afternoon and I looked forward to seeing him the next day.

My nightmare as a child was that the summer vacation will end and we will no more be together as a family as we had to go back to school. I remember my sleepless nights as the second half of the vacation started announcing that the end of the holiday is on its way. My only joy then was to be at home with my parents and siblings. Nothing more!

After finishing elementary school in Ramallah, I went to another boarding school in Beirut; leaving behind my siblings except for my eldest sister who went to university in Beirut and was able to see me at weekends. The fear of having summer vacations end grew more. During the summer holiday, I spent most of my time at home and hardly had the joy that young girls would have in going out. I was satisfied by being around family and by reading novels. My family and books were all what I needed.  I did not aspire to go out as my joy was by spending my time with my family.  I learnt to play backgammon, my father's favorite game so that I can play with him and spent time in the kitchen with my mother helping her. I did not want to lose any of those moments without my family.

Fear of separation from my family stayed with me as I went to university and when I started working in Bahrain. Sometimes our destiny is to stay separated. I hated saying goodbye to my parents and to my siblings and shed tears at every goodbye. Little did I know then that the hardest goodbye was few years ahead and I would be shedding tears all my life. I got married in Bahrain and continued to live away from my family. My tears never stopped when I said goodbye particularly to my parents. It took me several days to recover and settle after coming back home.

I had three wonderful kids and they and my husband were my new family. I could not imagine separating from them unless I had to. I was compelled to leave my two older kids with my family when I went to do my doctorate defense in the UK. My son was going to turn three that month and my daughter, two years in three months. I was eight months pregnant with my third and youngest child, so I guess I can say she came along with me. I cried my heart out leaving my two children during the five days I was away. I had sleepless nights and called them daily to hear their voices. As time went by and even after having my own family, I kept on crying while saying goodbyes to my parents and siblings. The day then came when my eldest child had to leave to university in the UK followed a year after by my second child and it was very painful. I knew of course that it is for their best but I could not help but suffer. The time came when my youngest daughter had to leave to university to join her siblings and keep the nest empty. I was so scared of this separation and she was worried about me. However, destiny had planned a forced goodbye for me, a separation unlike any other! My daughter died at the age of 18 in a car accident two weeks before going to university and I had to face the worst separation ever and the hardest goodbye!

Randah R Hamadeh

Math Portfolio (1)

Samar Al Ansari 

Grade 11.5

1.

a)   The pink graph represents the base function, y=x².  It is a quadratic function with a vertex of (0,0).

b) The black graph represents the function y=x²+3.  It is similar to the first graph, but it is shifted vertically three units upwards.  It has a vertex of (0,3).

c) The red graph represents the function y=x²-2.  It is similar to the first graph, but it is shifted vertically two units downwards.  It has a vertex of (0, -2).

The pattern I have noticed is that they all have the same basic parabola shapes, but is their vertical shifts which differentiate them. The general formula of quadratic equations is y=x²+k.  If k is positive, then the graph is shifted vertically upwards.  If k is negative, then the graph is shifted vertically downwards.  In the first function, y=x², k=0, so there is no shifted and the graph is located on the origin.  But in the second function, y=x²+3, k=3, a positive number, thus it is shifted vertically upwards three units.  In the third function, y=x²-2, k=-2,, thus it is shifted vertically downwards two units.

2.

a)     The pink graph represents the function, y=x².  The vertex of this graph is unchangeable, because nothing is altering the y-values or the x-values (nothing is being added/subtracted/multiplied to/divided by the x/y values).  It is the standard graph of a parabola, and is used as a comparison for the other functions.

b)    The black graph represents the function y=(x-2) ².  This function is shifted 2 units horizontally to the right, because the x-value is being altered by the opposite of the number added on to it within the parenthesis.  The vertex is changed from (0,0) to (2,0), because the vertex is shifted two units to the right since the equation is in the second degree and the opposite of the number within the parenthesis is the number changing the vertex.

c)     The red graph represents the function y=(x+3) ².  This function is shifted three units horizontally to the left.  The vertex is changed from (0,0) to (-3,0).

The pattern in these graphs is once again the same basic parabola shape, but their difference lays in the x-values; they are shifted horizontally either to the left or to the right.  It is obvious now that all equations rely on the formula y-k=a(x-h) ², the vertex being (h,k).  If h=positive, then the graph is shifted horizontally to the right, if h=negative then it is shifted horizontally to the left.  The same goes for k, if it is positive, the graph is shifted vertically upwards, if it is negative, it is shifted vertically downwards.  Thus in the second function, y=(x-2) ², the vertex will be (2,0) because the vertex is derived from this equation y-0=(x-2) ².  Thus the vertex of the third function y=(x+3) ² has a vertex of (-3,0) based on the formula.

3. The vertex would be (4,5) because the equation for this particular vertex is       y-5=(x-4) ², and the general formula is y-k=(x-h) ².  To get the vertex out of the equation you have to take the opposites of k and h in the formula to get (h,k). The first part is to take the 5 or k to the other side so you now get y-5=(x-4)^2. Then according to the formula you get a vertex of (4,5).

4.

a) x²-10x+25 in the form (x-h) ² is:

x²-10x+25=0                                                         work on the side: (10/2) ²=25

x²-10x=-25

x²-10x+25=-25+25

(x-5) ²=0

b) x²-10x+32 in the form (x-h) ² +g is:

x²-10x+32=0

x²-10x=-32                                   

x²-10x+25=-32+25

(x-5) ²=-7

(x-5) ² + 7=0

c)     Examples:

* 2x²+8x+1=0                                                       work on the side: (4/2) ²=4

2x²+8x=-1

2(x²+4x)=-1

2(x²+4x+4)=-1+8

2(x+2)²=-7

(x+2) ²=-7/2

(x+2) ² + 7/2

* 6x²+x-12=0                                                        work on the side: (1/6 /2)²=0.007

6x² + x=12

6(x² + 1/6x)=12

6( x² + 1/6x + 0.007)=12+0.042

6( x+ 0.083) ²= 12.42

(x+0.0083) ²=2.07

(x+0.0083) ² -2.07

* 9x²-36x-45=0                                                     work on the side: (4/2) ²=4

9x²-36x=45

9(x²-4x)=45

9(x²-4x+4)=45 +36

9(x-2) ²=81

(x-2) ²=9

(x-2) ²-9

d)    Steps:

1-    If c isn’t already on the other side, take it to the other side of the equation.

2-    If a≠1, set it equal to 1 by factoring it out.

3-    Divide b by 2, and square the answer you get.

4-    Add that to both sides of the equation.  Note: if a≠1 at the beginning, then when you add the answer of step 3 to the right side of the equation multiply it by the value of a at the beginning.

5-    Factor the left side of the equation.

6-    If a≠1 at the beginning, divide the right side of the equation by the value of a at the beginning.

7-    Add the right side of the equation to the left side of the equation in order for the answer to be in the (x-h) ² +g form.

5. The graph of (x-h) ² +g would be a parabola that is opened upwards, because it is a quadratic function.  There isn’t a negative sign outside of the brackets, thus the graph is not reflected on the x-axis.  If there was a negative sign outside the brackets, then the shape would be that of a parabola opening downwards.

The vertex of the function can be found by rearranging the formula to y-g=(x+h) ², thus the vertex would be (h,g).  There will be both a vertical and horizontal shift, it will shift vertically upwards, and horizontally to the right.  The vertex was originally located at (0,0) but has been shifted to (h,g).

     6.  This can be applied to absolute value function as well, y-k= a|x-h|

The blue function is the base function, with a vertex of (0,0).  The green function represents the function of y=|x| -2, thus it is shifted 2 units vertically downwards, making its vertex (0,-2).  The black graph represents the function y=|x-2|, thus the graph is shifted 2 units vertically to the right, making its vertex (2,0).  The general formula of an absolute value function is y-k=a|x-h|.  Thus to find the vertex, you multiply h and k by -1 making it (h,k). Question 1-a) and the blue function above share the same vertex (0,0) as well as an axis of symmetry of x=0 which is the same as the blue graph above.  Also, 1-c) and the green graph above share the same vertex (0,-2).  The general formula for 1-c) is y=x² +k, the general formula for the green graph can be written as y=|x|+k, since h=0.  The graph of 2-b) shares the same vertex of the black graph above (2,0).  All the absolute graphs follow the v-shaped pattern and are differentiated based on the vertical or horizontal shifts. The general formula of 2-b) could be written as y=(x-h) ², and the general formula of the black graph can be written as y=|x-h|, since k=0.  Although quadratic functions create a parabola graph and absolute value functions create v-shaped graphs they both have axis of symmetry and vertexes. The graph of a quadratic function is in the second degree while the graph of an absolute value function is in the first degree.  Both types of functions are symmetrical to the y-axis.  Also, their general formulas are similar because they share the same transformations.  The general formula of quadratic functions is y-k=a(x-h) ², while the general formula for absolute value functions is

y-k=a|x-h|.  

The graph above shows the functions of radical equations, the pink graph represents the function, y=√x, the black graph represents the function, y=√x  -2, and the red graph represents the function y=√(x-2 ).  The pink graph represents the base function, with a vertex of (0,0).  The black graph is shifted two units down, shifting the vertex to (0,-2).  The red graph is shifted two units to the right, making its vertex (2,0).  Question 1)a and the pink function above share the same vertex (0,0), question 1)c and the black function above share the same vertex (0-2), and question 2)b and the red function above share the same vertex (2,0). Once again they all follow the same general pattern which is the shape of the graph above, but differ in either terms of x-values or y-values based on what type of shifts are implemented. The general formula of graphs containing radical function is y-k=√(x-h), which is very similar to the general formula of a quadratic function.