## PedroSoccer 3 years ago 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

1. PedroSoccer

2. Barcasucks123

no

3. ipm1988

cannot see the graph

4. PedroSoccer

That was not necessary @Barcasucks123.

5. PedroSoccer

@ipm1988 I'll work on it :)

6. PedroSoccer

hmm. I'm not sure why it wouldn't show up... :\$

7. PedroSoccer

@ipm1988 does it still not show up?

8. ipm1988

nope just use print screen and post a picture

9. PedroSoccer

hmm my computer won't let me. I used control + p and everything

10. PedroSoccer

If I click on the link, it shows up for me. maybe my computer is having problems or something. Thanks though! :)

11. PedroSoccer

Can you try copying the link and opening it in a new tab though? Just to try?

12. AccessDenied

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.

13. PedroSoccer

Oh, that might be it. Thanks! :)

14. AccessDenied

If possible, you could try drawing what the graph looks like here, including any details like intercepts. :)

15. PedroSoccer

Okay, I'll give it a shot :)

16. PedroSoccer

|dw:1349652754929:dw|

17. PedroSoccer

Does that give enough info?

18. AccessDenied

Hmm... so those intercepts are (-2, 0), (2, 0), and (0, -4)?

19. PedroSoccer

Yes :)

20. AccessDenied

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? :)

21. PedroSoccer

Yes! Thanks so much! :)

22. AccessDenied

You're welcome!