Which method can you use to prove these triangles congruent?
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I'm going to call the function for the graph of A, A(x) and the function for the graph of B, B(x),and the function for the graph of C, C(x)
we know that when A(x)=0 (and x>0), C(x)=B(x) at that value of x
also, when B(x)=0, A(x)=C(x) at that value of x
we need to find the intersection points
a quadratic function can be written in form y=a(x-b)(x-c) where the roots or zeroes are at x=b and x=c and a is a constant. since we can only find 2 points, we will assume a=1 since in the other quadratic function, a=1 as well
let's use our 1st discovery first
when A(x)=0, (and x>0), C(x)=B(x)
find zeroes of A(x)
find where it equals 0
, factor, what 2 numbers mulitly to get -3 and add to get 2
0=(x-1)(x+3)
set equal to 0
0=x-1, x=1
0=x+3, x=-3
we are looking for positive root, so at x=1, C(x)=B(x)
meaining C(1)=B(1), 3(1)+3=B(1), 3+3=B(1), B(1)=6, that's one point
2nd thing
when B(x)=0, A(x)=C(x)
solving A(x)=C(x) for x<0 (since it's to the left of y axis)
factor, what 2 numbers multiply to get -6 and add to get -1
(x-3)(x+2)=0
set equal to 0
x-3=0, x=3
x+2=0, x=-2
we want x<0 so at x=-2
find y value, C(3)=3(-2)+3=-6+3=-3
we know that B(x) has a zero at (-3,0)
we also know that B(x) passes through the point (1,6)
if a quadratic function has zeroes at m and n, then it can be factored into form y=a(x-m)(x-n) where a is a constant, we assume a=1 because otherwise we can't find the function since we only know 2 points
we know that one zero is at -3,
y=(x-(-3))(x-n)
y=(x+3)(x-n)
subsitute the other point (1,6) to find n
6=(1+3)(1-n)
6=(4)(1-n)
6=4-4n
2=-4n
n=-1/2
so the equation is
or
the only problem is that in the picture, the graph of B crosses the x axis at a positive and negative value but the equation I have crosses the x axis at 2 negative values
but without further information, I can't solve for B