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OpenStudy (anonymous):
Can someone check my answer?
Find the equation for the graph.
The equation I got is y=2x^2
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OpenStudy (anonymous):
hartnn (hartnn):
in the graph, at x=1
y=4
thats not true according to your equation
hartnn (hartnn):
also, at x=0, y=2
hartnn (hartnn):
thats an exponentially growing function
so try to make a function where exponent is 'x'
OpenStudy (anonymous):
oh ok. Could you show me step by step? I learn that way a lot easier.
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hartnn (hartnn):
when
x= 0, y=2
x=1, y= 4
x=2, y=8
...
x= -1, y=1
hartnn (hartnn):
that 2,4,8....sequence should remind you of \(2^n\) function
hartnn (hartnn):
but in that n starts from 1
here, when x=1, y =4
2^2 is 4
so exponent must be x+1
\(\Large 2^{x+1}\)
OpenStudy (anonymous):
oops I mean I got y=x^2
hartnn (hartnn):
no...thats not an exponential function either
\(\large y= 2^{x+1}\)
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OpenStudy (anonymous):
is y=2(2)^x an exponent?
OpenStudy (anonymous):
Sorry this is just confusing me haha.
hartnn (hartnn):
yes
\(y = 2^{x+1} = 2 \times 2^x\)
OpenStudy (anonymous):
Oh ok! so y=2(2)^x is the answer?
hartnn (hartnn):
yup :)
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