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 $y$ :

$y-1>0$

$y>1$

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