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anonymous
 3 years ago
ohkay so I REALLY need help.
anonymous
 3 years ago
ohkay so I REALLY need help.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x  h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what grade of math are you in?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Solving Quadratic Equations

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im doing a project thing. and I was going thru the steps then I go to that one and I stopped cause I didnt know what to do

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its the pic in the attatchment

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you have access to a digital camera, take a picture of a parabola existing in the real world. Upload that picture onto your computer's hard drive. If you don't have access to a digital camera, search the internet for an image of a parabola existing in the real world. Be sure to follow the guidelines and safety precautions for completing internet searches. In a new GeoGebra window, select the axes button or the grid button underneath the toolbar if you cannot see either in your GeoGebra window. Deactivate the Algebra window. Go to View in the Menu bar. Deselect Algebra. Confirm there is a check mark next to Input Bar, also under View in the Menu bar. Insert the image from your digital camera or internet search into GeoGebra using the Insert Image button . Remember that this tool might be under one of the other buttons. Select the tiny triangle in the bottom right corner of each button until you find it. Once it’s activated, insert the picture by selecting the coordinate grid. You may need to adjust the placement of your image on the screen. To do this, rightclick on the image and select Object Properties… Under the Position tab, enter the coordinates of two or three corners of your picture.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Select the point tool , then place a point at the vertex of the parabola in your image. Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x  h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar. See how closely the graphed parabola matches your image parabola. You may need to try different values of a to make the parabola fit. To change the values of the equation, rightclick on the parabola, choose Object Properties…, and change the equation under the Basic tab. Close the window. Once the graphed parabola matches as close as possible to the image, make note of the quadratic equation, both in general and standard form (y = ax2 + bx + c). When completed, the graphed parabola should lie on top of the parabola in your image. If necessary, rightclick on the image and go to the Color tab to make the image more transparent so the graphed parabola can be seen over the image.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Identify the vertex, axis of symmetry, domain and range of the graphed parabola by inspecting your graph. Find the x and yintercepts of the parabola using the "Intersect Two Objects" icon . Remember this tool may be hidden below other buttons. To find the xintercepts, select the parabola, then the xaxis. To find the yintercepts, select the parabola, then the yaxis. Find the discriminant and explain the best method of finding the xintercepts algebraically. Find the xintercepts, or solutions to the quadratic equation, algebraically using each of the following methods: factoring, the quadratic formula, and completing the square. If a method cannot be used, explain why. Show all work involved with each method. Write a paragraph or two describing the picture you took. Provide details such as where it was found, how you found it, and why you chose this parabola in your activity. Send the activity to your instructor with the following information: Title (1 point) Materials Used (1 point) Procedure (1 point) Data(20 points) Include: A screenshot of GeoGebra with the picture you uploaded in the background and the graphed parabola on top. This can be done using the menu at the top of the GeoGebra screen. Go to "File," "Export," "Graphics View to Clipboard." Rightclick in your document and choose "Paste." (2 points) The quadratic equation matching your picture in general form and in standard form. (2 points) Vertex, axis of symmetry, domain and range of the parabola. (4 points) The x and yintercepts found using GeoGebra. (2 points) A complete sentence explaining which method of solving would be best used based on the value of the discriminant. (2 points) The solutions, or xintercepts, of the parabola found algebraically by factoring, using the quadratic formula and completing the square. If one of these methods cannot be used, a complete sentence explaining why. (3 points) 12 paragraphs describing the parabolic image chosen. Provide details such as where it was found, how you found it, and why you chose this parabola in your activity. (5 points)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, a parabola can be defined by its x intercepts, called zeroes or roots. For what x values does your parabola cross the x axis at?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i cant tell that part, pictures not that detailed for me to see, dw:1365103947488:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0are those x values 2 and 8 on the picture you posted?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0you have 2 arches ... which one are you using?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if its the front one, im guessing its a 2.5 and a 12.5 but im not sure what you are looking at

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0...why? front one seems simpler; 3 and 13, while the back one might be iffier

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0lol, do you agree that 3 and 13 look about right for those x values?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0we will also want to know the value of x that is in the center of those, that will be our "line of symmetry".

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0what value is the center between 3 and 13?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if you cant find the middle between 3 and 13, then you are definantly not prepared to do quadratics ....

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0we want some value such that 3+n = 13n can you solve for n?
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