ASSIGNMENT 2: Explorations with the Graphs of Parabola

By Nikhat Parveen, UGA

Explorations of the Parabola

y = 2x2 +3x - 4

In this assignment we will explore the various forms of the graph 2x2 +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  2x2 +3x - 4 over x-axis to produce the graph concave down.

 

Let's explore the graph of the Parabola  y = 2x2 +3x - 4.

 

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

Substituting 'x' with (x - 4) in the quadratic equation y = 2x2 +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 = 2x2 -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 = 2x2 +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 = 2x2 +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 = 2x2 +3x - 4

 y = 2(x + 4)2 + 3(x + 4) - 4

y = 2x2 + 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 = 2x2 +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 x2) from the original graph and the vertex of the graph is (-3/4, -5.125). (we know this from the graph of y = 2x2 +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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