Mathematics
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OpenStudy (gabylovesyou):
Based on the graph of y = x2 − 2x − 3, what is the x coordinate of the positive x-intercept?
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OpenStudy (gabylovesyou):
@dumbcow
OpenStudy (aravindg):
The x intercepts will be the roots of the given quadratic
OpenStudy (mertsj):
Find the x intercepts by replacing y with 0
OpenStudy (aravindg):
Out of the 2 roots .The one which is positive will be the x intercept you are looking for.
OpenStudy (aravindg):
Got it?
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OpenStudy (gabylovesyou):
0 = x^2 - 2x - 3
what do i do after this @Mertsj
OpenStudy (gabylovesyou):
@AravindG how do i solve it ?
OpenStudy (mertsj):
I think you are supposed to graph it
OpenStudy (aravindg):
yes^^
OpenStudy (mertsj):
The problem says...based on the graph
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OpenStudy (mertsj):
So do you know how to find the vertex?
OpenStudy (aravindg):
Or you just factorize :)
OpenStudy (gabylovesyou):
vertex is x = -b/2a right ?
OpenStudy (mertsj):
yes. That is the x coordinate of the vertex.
OpenStudy (gabylovesyou):
ok i will solve it now hold on
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OpenStudy (mertsj):
ok
OpenStudy (gabylovesyou):
is it x = -(-2) / 2(1) ??
OpenStudy (mertsj):
Yes. so x = 1
OpenStudy (gabylovesyou):
now ill solve for y hold on
OpenStudy (gabylovesyou):
y = -4
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OpenStudy (mertsj):
Graph that point. (1,-4)
OpenStudy (gabylovesyou):
now what
OpenStudy (mertsj):
Replace x with 0 and find y
OpenStudy (gabylovesyou):
but to what equation ? lol
OpenStudy (mertsj):
The only equation we have. y=x^2-2x-3
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OpenStudy (mertsj):
|dw:1369505987266:dw|
OpenStudy (gabylovesyou):
y = -3
OpenStudy (gabylovesyou):
right @Mertsj
OpenStudy (mertsj):
Then use symmetry to find the corresponding point to the right of the vertex...as shown in the drawing.
OpenStudy (mertsj):
Now let x = -1 and find two more points.
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OpenStudy (gabylovesyou):
y = 0
OpenStudy (mertsj):
|dw:1369506239595:dw|