anonymous
  • anonymous
What is the equation of the following graph?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
https://static.k12.com/eli/bb/179/1_17019/2_7354_8_17079/eec3393a134dfa3a7ae7a47f518e5a3db72f35ef/media/5782e5b6e9454cd7f6f7db8cb49fb6219a6e31c9/mediaasset_669401_1.bmp
anonymous
  • anonymous
please help me!
anonymous
  • anonymous
This is just like the other one, except now you have to use the graph to pick out the values. What is the vertex?

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anonymous
  • anonymous
(3,-2)?
anonymous
  • anonymous
yes. and the roots?
anonymous
  • anonymous
i'm not sure what the roots are...
anonymous
  • anonymous
the roots are where the graph crosses the x-axis. For example|dw:1434816829118:dw|
anonymous
  • anonymous
What is the x where your parabola crosses?
anonymous
  • anonymous
1 and 5?
anonymous
  • anonymous
yes. Now try plugging the roots into the equation. Remember the form is \[y = a(x-p)(x-q)\] Where p and q are the roots
anonymous
  • anonymous
-2=a(3-4)(3-3)?
anonymous
  • anonymous
not quite. We're not using the vertex yet or plugging in for x and y... yet. We have to set up the equation first. The root are 1 and 5, so plug in 1 and 5 for p and q to get y = a(x - 1)(x - 5) *It doesn't matter which is p and which is q*
anonymous
  • anonymous
NOW you plug in the vertex to solve for a
anonymous
  • anonymous
oh, okay. let me try.
anonymous
  • anonymous
-2=a(3-1)(3-5)
anonymous
  • anonymous
yes, now solve it for a
anonymous
  • anonymous
thanks. hold on.
anonymous
  • anonymous
a=1/2?
anonymous
  • anonymous
yes. can I see your final equation to make sure you got the format right?
anonymous
  • anonymous
sure.-2=a(3-1)(3-5) a=1/2
anonymous
  • anonymous
The a, p, and q are constants that we used properties to find. They are the same for every point on the parabola. The x and y are variables and change for every point, so we need them to stay. So the equation is y = ½(x -1)(x - 5)
anonymous
  • anonymous
so i solve this equation?
anonymous
  • anonymous
no, that's the answer
anonymous
  • anonymous
thanks again man, this helped.
anonymous
  • anonymous
you're welcome

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