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OpenStudy (misty1212):
HI!!
OpenStudy (twizttiez):
Hello!
OpenStudy (misty1212):
\[y=kx^2\] we need \(k\)
OpenStudy (twizttiez):
Would k be 32?
OpenStudy (twizttiez):
Or no y is 32
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OpenStudy (misty1212):
no dear
put \(x=4,y=32\) and solve
\[32=k\times 4^2\] for \(k\)
OpenStudy (misty1212):
\[32=16k\\k=?\]
OpenStudy (twizttiez):
Ok soo k=2?
OpenStudy (misty1212):
yes, and the equation is
\[y=2x^2\]
OpenStudy (anonymous):
@TwiztTiez After misty finishes helping you, compare this to the previous question and look at the pattern, ok? =]
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OpenStudy (twizttiez):
Ok thanks
OpenStudy (misty1212):
\[\color\magenta\heartsuit\]
OpenStudy (twizttiez):
But wait now that i have k would i do 32=2x4^2??
OpenStudy (twizttiez):
@misty1212
OpenStudy (twizttiez):
??
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OpenStudy (twizttiez):
Anyone?
OpenStudy (twizttiez):
@Nnesha
OpenStudy (twizttiez):
Hello?
OpenStudy (twizttiez):
@pitamar is that what i do?
OpenStudy (twizttiez):
Hello!?
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OpenStudy (anonymous):
Sorry, I'm back. let me see
OpenStudy (twizttiez):
Ok
OpenStudy (anonymous):
The thing is that you got k out of 32 = k*4^2. that's why you know k=2.
The power this gives you is that now you know the equation is:
$$
y = 2 \cdot x^2
$$
And you can find the 'y' values for other points.
Take a point you see in the graphs, (say x=2 as we did in previous question), plug it in the equation and see what the 'y' value has to be.
Now check what graph is matches those (x,y) values.
OpenStudy (twizttiez):
So A?
OpenStudy (anonymous):
Yes
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OpenStudy (anonymous):
Now look at the previous question and see that it is pretty much the same and that you understand the idea. ok?