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alainabbyboo22

  • 3 years ago

Which equation has this graph?

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  1. alainabbyboo22
    • 3 years ago
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  2. alainabbyboo22
    • 3 years ago
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    Answer choices available: A. y = x2 - 6x + 5 B. y = -x2 - 6x + 5 C. y = -x2 - 6x - 5 D. y = x2 + 6x - 5

  3. heisenberg
    • 3 years ago
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    You said you tried plugging in some points from the graph in these equations? What points did you choose?

  4. alainabbyboo22
    • 3 years ago
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    Oh that was for a different equation. This one is a new one.

  5. heisenberg
    • 3 years ago
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    Ah, well same method should work. There are two points that are very easy to spot here: the "zeros" of the graph: (1,0) and (5,0) do you see that?

  6. alainabbyboo22
    • 3 years ago
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    Not exactly.. :/ I don't get this one.

  7. heisenberg
    • 3 years ago
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    So can you tell me any point on that graph? A point is an x and y coordinate.

  8. alainabbyboo22
    • 3 years ago
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    1 and 3? I don't know. :(

  9. heisenberg
    • 3 years ago
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    A point looks like this: (x, y). These numbers correspond to points on graphs. Do you see how (1,0) lies on that graph? Do you see how (0,0) does not?

  10. alainabbyboo22
    • 3 years ago
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    Yeah I suppose.

  11. heisenberg
    • 3 years ago
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    |dw:1327095609111:dw| like that^

  12. alainabbyboo22
    • 3 years ago
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    Alright, then what's next?

  13. heisenberg
    • 3 years ago
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    The graph "passes through" (1,0) but not (0,0)

  14. heisenberg
    • 3 years ago
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    So we know (1,0) is a point on this graph. So let's try and plug x = 1 and y = 0 into those 4 choices and see which one works.

  15. heisenberg
    • 3 years ago
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    A. y = x2 - 6x + 5 B. y = -x2 - 6x + 5 C. y = -x2 - 6x - 5 D. y = x2 + 6x - 5 Can you plug x = 1 and y = 0 into any of these?

  16. heisenberg
    • 3 years ago
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    I'll do the first one for you: y = x^2 - 6x + 5 if x = 1, and y = 0 we get: 0 = (1)^2 - 6(1) + 5 0 = 1 - 6 + 5 0 = -5 + 5 0 = 0 So this is true for equation A. Equation A is a possible answer.

  17. heisenberg
    • 3 years ago
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    Are you stuck/confused?

  18. alainabbyboo22
    • 3 years ago
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    Hold on I'm trying to figure them out.

  19. heisenberg
    • 3 years ago
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    !!! Nice :)

  20. alainabbyboo22
    • 3 years ago
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    For the B, I solved the equation and got the answer 6. Where do I go from there for B?

  21. heisenberg
    • 3 years ago
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    You should have gotten an equation for equation B: something like: ? = ?

  22. heisenberg
    • 3 years ago
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    not just a single number, but an equation.

  23. alainabbyboo22
    • 3 years ago
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    Well I'm saying where do I go after I solve for that? To get to ?=?

  24. heisenberg
    • 3 years ago
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    So let's take a look at B: y = - x^2 - 6x + 5 if x = 1, y = 0, plug them in the equation: 0 = - (1^2) - 6(1) + 5 0 = -1 - 6 + 5 0 = -2 but this is not true! 0 does not equal -2, do you agree?

  25. alainabbyboo22
    • 3 years ago
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    I agree! So basically, once I have plugged in everything, I get rid of the x's and just put those together and I will find out what 0 equals?

  26. alainabbyboo22
    • 3 years ago
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    A will be my answer because that is the only one that equals 0!

  27. heisenberg
    • 3 years ago
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    You want to find the equation where the points (x=1,y=0) and also (x=5,y=0) satisfy the equation.

  28. heisenberg
    • 3 years ago
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    I think A is right as well!

  29. alainabbyboo22
    • 3 years ago
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    Thank you heisenberg! :)

  30. heisenberg
    • 3 years ago
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    Np :). Hope you understood it a little bit. Basically, this was our process: 1. Find a point that lies on the graph 2. Check each equation to see if this point's x and y result in a *true* equation. Good luck :)

  31. alainabbyboo22
    • 3 years ago
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    Thanks!! :)

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