**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.

- 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.

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