Which equation matches the graph shown?
(will attach graph and answer choices in a sec!)
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OpenStudy (anonymous):
GRAPH
OpenStudy (anonymous):
Answer choices A,B,C,D top to bottom
OpenStudy (anonymous):
roots
OpenStudy (anonymous):
then y-int
OpenStudy (anonymous):
i think B and D are out because that would mean the graph would be going upwards and the graph is going downwards... am i right??? the answer choices are between A and C??
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OpenStudy (anonymous):
y int=2
roots= (-2,0) and (1,0)
??? idk is that right??
OpenStudy (anonymous):
yes thats true
OpenStudy (anonymous):
yes then use in root form
OpenStudy (anonymous):
nice elimenation teqnique
OpenStudy (anonymous):
so how do i solve?? whats the root form again?? ;O
is it
y=a(x-h)^2+k
???is that it?
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OpenStudy (anonymous):
oh and thanks! :D
OpenStudy (anonymous):
\[y=a(x-x _{1})(x-x _{2})\]
OpenStudy (anonymous):
ohhh okay so what do i plug in again?
OpenStudy (anonymous):
i always forget this one :/
OpenStudy (anonymous):
first plug roots in (planting)
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OpenStudy (anonymous):
okay which ones do i plug? can u pls show me?? I'm a bit confused :/
OpenStudy (anonymous):
both roots
OpenStudy (anonymous):
x intercepts
OpenStudy (anonymous):
i do them separately? like i do it twice? once for each root?
OpenStudy (anonymous):
no same time
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OpenStudy (anonymous):
\[y=a(x+2)(x-1)\]
OpenStudy (anonymous):
did u just plug them in?
OpenStudy (anonymous):
and what do i do now then?
OpenStudy (anonymous):
plug y-int
OpenStudy (anonymous):
into the y value part?
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OpenStudy (anonymous):
@Jonask what do i do now>????
OpenStudy (anonymous):
\[2=a(0-1)(0+2)\]
OpenStudy (anonymous):
so 2=a(-1)(2)
2=a(-2)
-1=a
??
is that right?
OpenStudy (anonymous):
@jonask is that right?? if so, what do i do next??
OpenStudy (anonymous):
yes,you are done
\[y=-1(x-1)(x+2)\]
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