anonymous
  • anonymous
Which equation has this graph?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
Answer choices available: A. y = x2 - 6x + 5 B. y = -x2 - 6x + 5 C. y = -x2 - 6x - 5 D. y = x2 + 6x - 5
heisenberg
  • heisenberg
You said you tried plugging in some points from the graph in these equations? What points did you choose?

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anonymous
  • anonymous
Oh that was for a different equation. This one is a new one.
heisenberg
  • heisenberg
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?
anonymous
  • anonymous
Not exactly.. :/ I don't get this one.
heisenberg
  • heisenberg
So can you tell me any point on that graph? A point is an x and y coordinate.
anonymous
  • anonymous
1 and 3? I don't know. :(
heisenberg
  • heisenberg
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?
anonymous
  • anonymous
Yeah I suppose.
heisenberg
  • heisenberg
|dw:1327095609111:dw| like that^
anonymous
  • anonymous
Alright, then what's next?
heisenberg
  • heisenberg
The graph "passes through" (1,0) but not (0,0)
heisenberg
  • heisenberg
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.
heisenberg
  • heisenberg
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?
heisenberg
  • heisenberg
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.
heisenberg
  • heisenberg
Are you stuck/confused?
anonymous
  • anonymous
Hold on I'm trying to figure them out.
heisenberg
  • heisenberg
!!! Nice :)
anonymous
  • anonymous
For the B, I solved the equation and got the answer 6. Where do I go from there for B?
heisenberg
  • heisenberg
You should have gotten an equation for equation B: something like: ? = ?
heisenberg
  • heisenberg
not just a single number, but an equation.
anonymous
  • anonymous
Well I'm saying where do I go after I solve for that? To get to ?=?
heisenberg
  • heisenberg
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?
anonymous
  • anonymous
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?
anonymous
  • anonymous
A will be my answer because that is the only one that equals 0!
heisenberg
  • heisenberg
You want to find the equation where the points (x=1,y=0) and also (x=5,y=0) satisfy the equation.
heisenberg
  • heisenberg
I think A is right as well!
anonymous
  • anonymous
Thank you heisenberg! :)
heisenberg
  • heisenberg
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 :)
anonymous
  • anonymous
Thanks!! :)

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