anonymous
  • anonymous
How do I get the quadratic root equation numbers from a graph?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
well what does it mean to have the root of an equation? it means you have the 0s right?
anonymous
  • anonymous
i know how to balance, but how do I get the ax2+bx+c #s from it?
anonymous
  • anonymous
well first thing you want to do is find where it intersects the axis correct? where it intersects is going to give you your roots. so you will have something like (x+5)(x-3)

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anonymous
  • anonymous
then you can foil that guy out
anonymous
  • anonymous
oh, the points on the x-axis incept are the roots?
anonymous
  • anonymous
well isnt that how we defined the quadratic equation? the quadratic equation is set up such that your solving for the left hand side to equal 0
anonymous
  • anonymous
so if i have a graph and the two parabola intercepts are 2 and -2, then are those the roots, or are those the x +2 and x-2, or do they fit into the ax2+bx+c?
anonymous
  • anonymous
on the x-axis
anonymous
  • anonymous
oh you have multiple graphs?
anonymous
  • anonymous
no i have one graph, it's a parabola that intercepts the x-axis twice
anonymous
  • anonymous
whats the equation of a parabola? the general syntax? that should help you quite a bit if you analyze what happens with the parabola equation
anonymous
  • anonymous
i just don't know the equation for turning graph data into the qudratic root formula
anonymous
  • anonymous
ok well think about (x+2)(x-2)
anonymous
  • anonymous
that goes to x^2 - 4 correct?
anonymous
  • anonymous
your graph intercepts the y axis at -4 im guessing?
anonymous
  • anonymous
well.. you can change where it intercepts the y axis
anonymous
  • anonymous
is that "c", where it intercepts?
anonymous
  • anonymous
idk the equation for the parabola im justi thinking of how the graph works out intuitevely
anonymous
  • anonymous
i mean i have the equation. but think about it like this. (x^2 - 4)/c
anonymous
  • anonymous
the c is going to effect the 'width' of the graph
anonymous
  • anonymous
the x and y intercepts are still at 2 and -2
anonymous
  • anonymous
those are both the x intercepts the y-one is -4
anonymous
  • anonymous
but the c is going to change where it intercepts the y axis. you can solve for what c has to be though for the y intercept.
anonymous
  • anonymous
y = (x^2-4)/C
anonymous
  • anonymous
Set x to 0 and y to -4
anonymous
  • anonymous
solve for C
anonymous
  • anonymous
I know c=y-intercept, right? do i still use ax2... etc etc or do i use y=ax2 + c and how do i get "a"
anonymous
  • anonymous
well in my example my 'a' would be 1/C
anonymous
  • anonymous
and your 'c' would be my 4/C
anonymous
  • anonymous
so y = 1X^2-4 where a = 1 and C = -4...
anonymous
  • anonymous
X intercepts are the roots?

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