**ASSIGNMENT 2: Explorations with the Graphs
of Parabola**

**By Nikhat Parveen,
UGA**

Explorations of the Parabola

**y = 2x ^{2} +3x - 4**

In this assignment we will explore the various
forms of the graph 2x^{2} +3x - 4 by replacing x with (x - 4) and moving
the graphs in different quadrants. In short, we will investigate the various
positioning of the graph.

We will do this in following steps:

i. Replacing x with (x - 4)

ii Replacing x with (x + 4)

iii Reflecting 2x^{2} +3x - 4 over
x-axis to produce the graph concave down.

Let's explore the graph of the Parabola
**y = 2x ^{2} +3x - 4. **

The graph has the vertex (-0.75,-5.125).

Substituting 'x' with (x - 4) in the quadratic
equation y = 2x^{2} +3x - 4^{ }we get the horizontal translation
as shown in the following graph.

y = 2(x -4)^{2 }+ 3(x - 4) - 4

Expanding and simplifying we get,

y =^{ }2x^{2} -13x + 16

The graph shows a horizontal translation of 4
units to the right and we can be certain about this as the vertex was (0.75,
-5.125) for the previous graph. The vertex for this graph is (3.25,
-5.125).Comparing with the standard equation **y = a(x - h) ^{2 }+b(x-h)+
k**, where (h, k) is the vertex, we can be sure that the translation was as
expected and the vertex can be verified by completing the square.

The following graphs shows a horizontal
translations to the right with different substitutions of of 'x' with (x - 3),(x
- 2), (x - 1) in the Parabola y = 2x^{2} +3x - 4

The graph shows different vertex with different x -values but no change in y-values. This shows that any translation of (x-a), where 'a' is any positive integer, the graph will move horizontally to the right with change in the x-values but no change in 'y', ie.,the vertical translation will remain the same.

To move
the graph y = 2x^{2} +3x - 4
into the 2nd quadrant, we
must move the graph to the left. When we replaced x with (x-4), the graph was
translated to the right 4 units, so to move it to the left, we should replace x
with (x + c), where c is a positive value

Std.Equation:** **y = a(x + h)^{2 }+
b(x + h)+ k,

Let's substitute (x + 4) in the equation y = 2x^{2}
+3x - 4

y = 2(x + 4)^{2 }+ 3(x + 4) - 4

y =^{ }2x^{2} + 19x + 40

The graph moved to the second quadrant, but note the entire parabola does not lie in the second quadrant. It is partially in the third and if we trace the parabola it lies in the first quadrant as well.

Here the Parabola intersects the y-axis at its y-intercept.

So, let's change both the values 'h' as well as 'k' in order to move the graph completely into the second quadrant;

y = 2(x + 4)^{2 }+ 3(x + 4) + 4

What do we notice?? The graph moved to the second
quadrant by changing the value of k from -4 to +4.The vertical shift upward is 8
units with respect to the original graph and with respect to the graph
y = 2(x + 4)^{2 }+ 3(x + 4) - 4 the axis of symmetry is
the same h = -4.75, what changed is the y-value from -5.125 to 2.875, a vertical
upward shift of 8 units.

Graphs of horizontal translation on either side....

How can we produce a graph which is concave down that shares the same vertex as our original?

Our original graph is y =
2x^{2} +3x - 4 and to produce a graph that is concave down, simply means
we need to reflect the graph over x-axis. The standard form of the parabola with
intercept form is

y = a(x - h)^{2} +k, where (h, k) is the
vertex. For concave down we need to make 'a' negative.

y = -a(x - h)^{2} +k. Now, first we
need to convert the original graph to the intercept form. We know that a = 2
(leading coefficient of x^{2}) from the original graph and the vertex of
the graph is (-3/4, -5.125). (we know this from the graph of y = 2x^{2}
+3x - 4).

So substituting in the original equation we get:

y = 2(x
+3/4)^{2} -5.125

and to reflect this graph over x- axis to produce
concave down we have to add minus sign to the leading coefficient 2 . So, the
final equation is y = -2(x +3/4)^{2} -5.125

We were right!! The graph is concave down and has the same vertex as the original graph

(-0.75, -5.125). So, **y = -2(x + 3/4) ^{2}
- 5.125** is the equation of the graph that lies in the second quadrant.

By varying the value of 'a', do we get a different graph??

Yes, we do get different graphs but with the same vertex and different y-intercepts.

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