How do you find the domain and range of #y=2x^2+1#?

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Answer 1 · 2017-04-27T00:22:59

Domain: All real numbers
Range: #y>1#

Normally, #x# or #y# will be limited if there are fractions or square roots.

#x# is not limited to any numbers. That means the domain is all real numbers.

Solve for #y# to figure out the range:

#y-1=2x^2#

#(y-1)/2=x^2#

#sqrt((y-1)/2)=x#

#y# is in a square root this time and we need to figure out the limitations. A square root cannot have negative numbers. So:

#(y-1)/2>0#

Solve for #y#:

#y-1>0#

#y>1#