# Math 10 Ib Porfolio

Math 10C Pre-IB Portfolio Assignment Type 1 Investigating the Quadratic Function A quadratic function is one where the highest exponent of the independent variable is 2.The quadratic function can be written in the general form of, where a, b, and c are real numbers.However, the quadratic function can also be written in the standard form of , which is sometimes more preferred, where p and q are the x and y coordinates of the vertex, respectively.

The purpose of this task is to investigate the graph of a quadratic equation, the parabola, when the equation is written in the form.

**Math 10 Ib Porfolio**

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By analyzing p and q we can determine the vertex of the graph. Also, by increasing or decreasing the values of p or q, we can translate the parabola vertically and/or horizontally. First, if we look at the functions y=x2 ,y=x2+3, y=x2-2 we know that all 3 are in general form. To convert general to standard form you will need to use the process called “completing the square” which goes as following: Ex. y=ax+bx+c

Now if we convert the three functions mentioned above, in standard form respectively they are y=x020, y=x02+3 and y=x02-2. Now if we were to graph these points, either the standard or general form would work. y=x2 , y=x020 y=x2+3 , y=x02+3 y=x2-2 , y=x02-2 Other examples of these types of graphs could be anything along the line of. An example of a parabola in the form of y=x2q with either a positive or negative q value could be y=x2+5 and y=x2-4. When we graph the two equations they are as following: y=x2+5 , y=x02+5 y=x2-4 , y=x02-4

The reason I choose to convert them to standard form was to look for the vertex. By looking at the graph and the standard form of the equations we can conclude that the vertex of the graphs are (0,0) , (0,3) , (0,-2). In standard form you can also find out many other things. For example, the domain and range, the axis of symmetry. All three graphs are graphed in the same screen for comparison. From the above graph we notice that the graphs of y=x2+3 and y=x2-2 have been shifted vertically, either up or down by the q units, where the q is the number that follows the x2.

By looking at the following graph, we can generalize the following: The graph of is the graph of, vertically translated q units. If q is positive, then the shift is upwards. Conversely, if q is negative, than the shift is downwards. If we look at the next following functions of y=x2 , y=x-22 , y=x+32, we can see that the functions are written in the standard form. To change standard form to general form you need to expand the function. To do this you need to do the following steps: Ex. y=x+52 When we convert the functions of y=x2 , y=x-22 , y=x+32, respectively they are y=x20x0, y=x2-4x+4, y=x2+6x+9.

When we graph the following functions, we get: y=x20x0, y=x2 y=x2-4x+4, y=x-22 y=x2+6x+9, y=x+32 Furthermore, if we change the values of p in the function we can translate the graph to either the right or left. An example of this could be y=x-82 or y=x+52. When graphed they should look as the following: y=x2-16x+64, y=x-82 y=x2+10x+25,y=x+52 For the first three graphs above, the vertex of them respectively are (0,0), (0,2), (0,-3). The vertex of the above graphs are (0,8), (0-5).

While looking at the graphs we notice that if the value of p is changed to either a positive or negative number it depends whether the graph is shifted horizontally on the x-axis. As done previously, all three graphs have been graphed in the same screen for comparison. From the above combined graph, we notice that the graphs of y=x-22, y=x+32 have been shifted horizontally, to either the left or right by p units, where p is the number that follows after x. By looking at the above graph, we can generalize the following: The graph of is the graph of, horizontally translated p units.

If p is positive, the shift is to the right. On the other hand, if p is negative, the shift is to the left. Given the trends noticed when changing p and q, we can predict the vertex of the graph. The graph ofis the graph of, vertically translated 5 units upward, and horizontally translated 4 units to the right. Since the vertex of the graph isis at (0,0), the vertex of the graph should be at (4,5). Alternatively, since we know that the vertex of the parabolic function of the form lies at (p,q), we can expect that the vertex of the graphis to be at (4,5) as p=4 and k=5.