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Mathematics 13 Online
OpenStudy (anonymous):

What is the equation of the graph below? https://vcsohio.brainhoney.com/Resource/22138800,AB4,0,B,4/Assets/71871_53b5a642/5_14g_01.gif y = -2(x - 3)2 + 1 y = -2(x + 3)2 + 1 y = 2(x - 3)2 + 1 y = 2(x + 3)2 + 1

OpenStudy (anonymous):

@Directrix @sidsiddhartha

OpenStudy (sidsiddhartha):

firstly from the graph we can say that it is 3 unit from x axis and 1 from y axis .ok?

OpenStudy (anonymous):

so (3,1) right?

OpenStudy (sidsiddhartha):

yes so my first step will be \[y=(x-3)^2+1\]

OpenStudy (anonymous):

what's the equation for it?

OpenStudy (sidsiddhartha):

if the vertex is (h,k) then \[y=(x-h)^2+k\\here\\h=3,k=1\] ok?

OpenStudy (anonymous):

ok, I get that part, so what's next.

OpenStudy (sidsiddhartha):

now tell me that the graph opens downward or upward?

OpenStudy (anonymous):

downward, so it is negative right?

OpenStudy (sidsiddhartha):

yup good :)

OpenStudy (sidsiddhartha):

so \[y=-2(x-3)^2+1\]

OpenStudy (anonymous):

how did you get the 2?

OpenStudy (anonymous):

@sidsiddhartha ?

OpenStudy (sidsiddhartha):

uhh its lagging

OpenStudy (sidsiddhartha):

there shount be a 2

OpenStudy (anonymous):

thanks!

OpenStudy (sidsiddhartha):

y = -(x - 3)^2 + 1 should be this

OpenStudy (anonymous):

I have another one I need help with. I'll post it and tag you

OpenStudy (sidsiddhartha):

ok :)

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