What type of quadratic equation is represented in the graph below? Not enough information Non-factorable Trinomial Perfect Square Trinomial Difference of Two Squares http://learn.flvs.net/webdav/assessment_images/educator_algebra1_v10/09_02_08.jpg

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What type of quadratic equation is represented in the graph below? Not enough information Non-factorable Trinomial Perfect Square Trinomial Difference of Two Squares http://learn.flvs.net/webdav/assessment_images/educator_algebra1_v10/09_02_08.jpg

Mathematics
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The graph is in the link. Please help? Thanks! :)
no
cannot see the graph

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Other answers:

That was not necessary @Barcasucks123.
@ipm1988 I'll work on it :)
hmm. I'm not sure why it wouldn't show up... :$
@ipm1988 does it still not show up?
nope just use print screen and post a picture
hmm my computer won't let me. I used control + p and everything
If I click on the link, it shows up for me. maybe my computer is having problems or something. Thanks though! :)
Can you try copying the link and opening it in a new tab though? Just to try?
The issue is probably that we don't have accounts / are not logged into a system there like you are. :P "The system is unable to access the cookie that was assigned to your computer upon login. The most likely cause of this problem is an improper access to the course via a bookmark, browser window that was left open, or an improper exit from the system.
Oh, that might be it. Thanks! :)
If possible, you could try drawing what the graph looks like here, including any details like intercepts. :)
Okay, I'll give it a shot :)
|dw:1349652754929:dw|
Does that give enough info?
Hmm... so those intercepts are (-2, 0), (2, 0), and (0, -4)?
Yes :)
Well, with the x-intercepts / our zeros of the graph, we can tell that x = 2 and x = -2 are solutions. That means in terms of a quadratic equation, we have k(x - a)(x - b) = 0: with 'a =2' and 'b = -2' (The extra k is because we could have an extra constant multiplied on there). y = k(x - 2)(x + 2) = 0 y = k(x^2 - 4) Then for the extra 'k', we just have to use (0, -4): -4 = k(0^2 - 4) -4 = k(-4) 1 = k So, plug that in, we simply get: y = x^2 - 4 Looks a lot like a difference of two squares, right? :)
Yes! Thanks so much! :)
You're welcome!

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