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What is the inverse of the given relation? Y=7x^2-3
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y=11
Actually no sorry are you in algebra one or two?
wait is 7 mulitplying
I have no clue they Gave me a whole bunch of radicals and exclamations in the answers or I would put them up
3=14x
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solve y add 3 divide by 14
A. Y=+ or - sqrt of x+3/7 stuff like this
y+3=7x^2 x^2 = (y+3)/7 x= sqrt((y+3)/7) so we say : y=sqrt((x+3)/7)
this is the inverse.. (we solved for x)
y=+/-sqrt((x+3)/7)
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just interchange y and x then solve for y!! y=sqrt((x+3)/7)
\[y=7x ^{2}-3\] \[y-3=7x ^{2}\] \[\sqrt{\frac{ y-3 }{ 7 }}=x ^{2}\] So answer will be \[\pm \sqrt{\frac{ x-3 }{ 7 }}=y\]
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