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well what does it mean to have the root of an equation? it means you have the 0s right?
i know how to balance, but how do I get the ax2+bx+c #s from it?
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)
then you can foil that guy out
oh, the points on the x-axis incept are the roots?
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
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?
on the x-axis
oh you have multiple graphs?
no i have one graph, it's a parabola that intercepts the x-axis twice
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
i just don't know the equation for turning graph data into the qudratic root formula
ok well think about (x+2)(x-2)
that goes to x^2 - 4 correct?
your graph intercepts the y axis at -4 im guessing?
well.. you can change where it intercepts the y axis
is that "c", where it intercepts?
idk the equation for the parabola im justi thinking of how the graph works out intuitevely
i mean i have the equation. but think about it like this. (x^2 - 4)/c
the c is going to effect the 'width' of the graph
the x and y intercepts are still at 2 and -2
those are both the x intercepts the y-one is -4
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.
y = (x^2-4)/C
Set x to 0 and y to -4
solve for C
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"
well in my example my 'a' would be 1/C
and your 'c' would be my 4/C
so y = 1X^2-4 where a = 1 and C = -4...
X intercepts are the roots?