Mathematics
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OpenStudy (anonymous):
help graphing quadradtic function y+X^2+5x+2=0
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OpenStudy (anonymous):
the vertex of the function is:\[\frac{ -b }{ 2a }\]
OpenStudy (anonymous):
so if you write your equation in terms of y you get y = -x^2-5x-2
OpenStudy (anonymous):
so what is the a term?
OpenStudy (anonymous):
can u tell me?
OpenStudy (anonymous):
-x
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OpenStudy (anonymous):
the x-coordinate of the vertex is -b/2a so we need to know the a nd be terms
OpenStudy (anonymous):
yah so a is -1 right?
OpenStudy (anonymous):
what is b?
OpenStudy (anonymous):
-5
OpenStudy (anonymous):
yup so the x-cooridnate of the vertex of the parabola is -b/2a which is?
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OpenStudy (anonymous):
-(-5)/2(-1)=?
OpenStudy (anonymous):
?
OpenStudy (anonymous):
2.5
OpenStudy (anonymous):
actually it is -2.5
OpenStudy (anonymous):
cuz its 5/2(-1) = -2.5
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OpenStudy (anonymous):
how
OpenStudy (anonymous):
-(-5)/2(-1)
OpenStudy (anonymous):
-(-5) is 5
OpenStudy (anonymous):
2(-1) is -2
OpenStudy (anonymous):
so it is 5/-2 which is -2.5
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OpenStudy (anonymous):
ok so the x-coordinate of the vertex is -2.5 so how do u think u would find the y-coordinate?
OpenStudy (anonymous):
ok i get it
OpenStudy (anonymous):
plug in -2.5 for x
OpenStudy (anonymous):
yup
OpenStudy (anonymous):
so what do you get?
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OpenStudy (anonymous):
hold on
OpenStudy (anonymous):
i got 4.25
OpenStudy (anonymous):
-(2.5)^2-5(2.5)-2
OpenStudy (anonymous):
tok i have to go
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OpenStudy (anonymous):
so after you find the vertex find the y-intercepts where x=0
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
so the vertex is (-2.5,4.25) so graph that
OpenStudy (anonymous):
so plug in x=0 to find at which y-coordinate it crosses the yaxis
OpenStudy (anonymous):
so can u tell me where that is?
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OpenStudy (anonymous):
its -2 right?
OpenStudy (anonymous):
if you plug in x=0 you get -2
OpenStudy (anonymous):
yup
OpenStudy (anonymous):
so it crosses the y-axis at x=-2
OpenStudy (anonymous):
i mean at y= -2
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OpenStudy (anonymous):
so graph that
OpenStudy (anonymous):
it would be 2 below on the yaxis
OpenStudy (anonymous):
i have to plot 3 other points
OpenStudy (anonymous):
ok find the x-intercept where y=0
OpenStudy (anonymous):
didnt learn that
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OpenStudy (anonymous):
just plug in some number for x and find the y-coordinate
OpenStudy (anonymous):
and that will be a point on the parabola
OpenStudy (anonymous):
then do that for 2 other points
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
then you just connect them with curves and then you are done
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OpenStudy (anonymous):
do you get it now?
OpenStudy (anonymous):
YES