PhET-Simulations-for-Mathematics/C2/Graphing-Quadratics/English
--Madhurig (talk) 13:40, 17 November 2023 (IST)Title of the script: Graphing Quadratics
Author: Shraddha Kodavade
Keywords: Phet simulation, graphing quadratics, polynomial, parabola, roots, Axis of symmetry, directrix, vertex, intercept, video tutorial.
| Visual Cue | Narration |
| Slide Number 1
Title Slide |
Welcome to this tutorial on Graphing Quadratics. |
| Slide Number 2
Learning Objectives |
In this tutorial, we will learn about:
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| Slide Number 3
System Requirements |
This tutorial is recorded using,
Windows 10-64-bit operating system Chrome version 101.0.49 |
| Slide Number 4
Pre-requisites |
To follow this tutorial,
the learner should be familiar with topics in basic mathematics. Please use the link below to access the tutorials on PhET Simulations. |
| Slide Number 5
Link for Phet Simulations https://phet.colorado.edu/en/simulations/graphing-quadratics |
Please use the given link to download the PhET simulation. |
| Slide Number 6
PhET Simulations |
In this tutorial, we will use,
Graphing Quadratics PhET Simulation. |
| Show the Downloads folder.
Double click the file. |
I have already downloaded Graphing Quadratics PhET simulation to my Downloads folder. |
| Only Narration | Let us begin. |
| Cursor on the interface. | This is the interface of Graphing Quadratics simulation. |
| Point to the screens. | The interface has 4 screens.
Explore, Standard Form, Vertex Form, Focus and Directrix. |
| Click on Explore screen | Let us click on the Explore screen. |
| Cursor in the screen. | This screen helps to understand how the parameters influence the shape of the parabola. |
| Point to the centre graphing region | In the centre of the screen, we can see a Cartesian plane.
An orange colored parabola is shown by default. |
| Point to the upper right corner | This graph is of the form y=ax2+bx+c.
The equation of the graph is given below it, with a=1, b=0, and c=0. |
| Click on Quadratic terms. | Click the green plus button on the Quadratic Terms box.
Click on the Equations check box. The graph gets annotated with the equation y=x2. |
| Click on the grey boxes below the graphing plot. | Below the Cartesian plane we have two gray boxes with probes.
These boxes show the coordinates of a point on the plane. |
| Drag and place the probe of the first box at (1,1).
Drag and place the probe of the second box at (2,4). |
Drag and place the probe of the first box at (1, 1).
Drag and place the probe of the second box at (2, 4). These points satisfy the equation y=x2. So they lie on the graph. |
| Move the point probe from (2,4) to (2,0). | If we move the probe from (2,4) to (2,0), the colour of the coordinates change to white.
This indicates that, point does not lie on the graph. |
| Drag the slider of a. | Drag the slider of a.
If 'a is greater than zero(a>0), the parabola opens upward. If 'a is less than zero(a<0), the parabola opens downward. If 'a is equal to zero(a=0), the parabola becomes horizontal line at y=0. |
| Click the Reset button. | Click on the Reset button. |
| Drag slider b in positive direction | Drag the slider of b in positive direction.
The graph shifts to the second and third quadrants. |
| Drag slider b in negative direction | Drag the slider of b in the negative direction.
Now the graph shifts to first and fourth quadrants. |
| Click on Reset option. | Click on the Reset option. |
| Drag the slider of c. | Drag the slider of c.
If c is greater than zero(c>0), the curve shifts vertically upwards to the first and second quadrants. If c is less than zero(c<0), the curve shifts vertically downwards to third and fourth quadrants. |
| Change the equation to y=x2+x+1
Click on quadratic terms |
Change the equation to y=x2+x+1.
Here a= 1, b= 1 and c= 1. Click on the Quadratic Terms and then click on Equations. |
| Click on y=ax2 | Click on y=ax2.
The equation of the blue graph is y=x2 . It is representative of only the quadratic term of the equation. The x and the constant terms over here vanish. |
| Click on y=bx. | Click on y=bx.
The equation of the green graph is y=x . It is representative of only the linear term of the equation. The x2 and constant terms vanish here. |
| Click on y=c. | Click on y=c.
The equation of the pink graph is y=1. It is representative of only the intercept of the equation. The x2 and x terms vanish here. |
| Click on the second screen -Standard Form | Now, let us explore the Standard Form screen.
This screen is similar to the previous one. Three new concepts are introduced here- Vertex, Axis of Symmetry and Roots. The default equation is y=x2. |
| Click the Vertex check box.
|
Click on the Vertex check box.
It is denoted by a purple dot. In this equation, the vertex is at (0, 0). Click on the Equations checkbox. |
| Click on Axis of symmetry | Click on the Axis of Symmetry check box.
It is a vertical line that divides the parabola into two equal halves. |
| Point to the purple dashed line | It is denoted by a purple dashed line.
For the equation y=x2, the axis of symmetry is the y-axis. This means that the parabola is symmetric with respect to y axis. |
| Click on Roots checkbox. | Click on the Roots checkbox.
It is denoted by two blue dots. Roots are the points of intersection where the parabola meets the x-axis. The nature of the roots may be real or imaginary. For this equation there is a single root at (0,0) |
| Point to the vertex.
Point to the line of symmetry Point to the roots |
Let us make the new equation y=x2+3x.
Here, the vertex is at (-1.5, -2.25). The line of symmetry is x= 1.5. There are two roots, (-3,0) and (0,0). |
| Point to NO REAL ROOTS. | Let us change the equation to y=x2+3x+3.
The vertex changes to (-1.5, 0.75). There are no real roots for this equation. This is because the parabola does not touch the x-axis. |
| Click the arrows of a, b and c to change the values. | Let us change the values of a, b and c and check the graph and its roots. |
| Click on the fourth screen - Focus and Directrix | Let us explore the fourth screen-Focus and Directrix.
On this screen we see a parabola with equation y=1/4p(x-h)2+k. By default the equation is y=1/4(2.0)(x-0)2+0, where p=2, h=0 and k=0. |
| Click on point on Parabola and equations. | On the Cartesian plane, an upward opening parabola is shown.
Vertex, Focus and Directrix check boxes are selected by default in the Options panel. |
| Click on Point on parabola and Equations
Point to the Vertex |
Click on Point on parabola and Equations check boxes.
Here, the vertex is at (0,0). The equation reduces to y=0.125x2. |
| Point to the green dot.
Point to the directrix. |
In this figure, the focus is denoted by a green dot at point (0, 2).
The directrix is denoted by green dashed line at y=-2. |
| Cursor on Point on the parabola. | The Point on the parabola denotes the distance from the directrix and the focus. |
| Drag the parameter p. | There are three parameters for this equation: p, h and k.
Drag the slider of parameter p. |
| Slide p>0. Stop at p=9.0
Point to focus (0,9) and directrix (y=-9) |
As we increase the value of p, the parabola becomes flat.
For p=9.0, the focus becomes (0,9) and the directrix becomes (y=-9). The new equation of parabola is y=0.028x2. |
| Slide p<0. Stop at p=-9.0
Point to focus (0,-9) and directrix (y=9)
|
As we decrease the value of p, the parabola becomes flat in opposite direction.
For p=-9.0, the focus becomes (0,-9) and the directrix becomes (y=9). The new equation of parabola is y=-0.028x2. The focus and directrix move farther away from the original graph. Let’s bring back the p value to 2. |
| Slide the parameter h.
Slide h and stop at h=6.0 Point to focus (6,2) and directrix (y=-2) Slide p. Stop at h=0.0 |
Drag the slider of parameter h.
For h>0, the parabola shifts horizontally in positive direction of the x-axis. For h<0, the parabola shifts horizontally in negative direction of the x-axis. For h=6.0, the new equation of the parabola is y=0.125(x-6)2. Notice that the vertex has shifted from (0,0) to (6,0). The focus is at (6,2) and the directrix at y=-2. Let us bring h value to 0. |
| Slide the parameter k.
Slide k and stop at k=-6.0 Point to focus (0,-4) and directrix (y=-8)
|
For k=-6.0 The new equation is y=0.125x2-6.
Notice the vertex has shifted from (0,0) to (0,-6). The focus is at (0,-4) and the directrix is at y=-8. |
| With this we have come to the end of this tutorial.
Let us summarise. | |
| Slide Number 7
Summary |
In this tutorial, we have learnt about:
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| Slide Number 8
Assignment |
As an assignment to this tutorial,
Explore the third screen - Vertex Form. Change the values of a, h and k and Check the graph and its roots. |
| Slide Number 9
About the Spoken Tutorial Project |
The video at the following link summarises the Spoken Tutorial project.
Please download and watch it. |
| Slide Number 10
Spoken Tutorial workshops |
The Spoken Tutorial Project team:
conducts workshops and gives certificates on passing online tests. For more details, please write to us. |
| Slide Number 11
Forum for specific questions: Do you have questions about THIS Spoken Tutorial? Please visit this site. Choose the minute and second where you have the question Explain your question briefly Someone from our team will answer them |
Please post your timed queries in this forum. |
| Slide Number 12
Acknowledgement |
The Spoken Tutorial project is funded by Ministry of Education, Govt. of India |
| Slide Number 13
Thank you
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This is Shraddha Kodavade a FOSSEE summer fellow 2022, IIT Bombay signing off.
Thanks for joining.
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