## Friday, February 22, 2013

### Math Portfolio (1)

Samar Al Ansari
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.

1. 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.