# State the domain and range of each function. Justify your answer. y= 2(x-3)^2 + 1

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### 2 Answers

Hi, super,

The domain of this function is "all real numbers". Many times, but not everytime, you would simply consider "the nature of the function". As in, with this function, you can plug in anything for x and get something out. It's not like y = 1/x, where x can't be 0, because you would divide by 0. Here, you don't have to worry anything about that. The domain is "all real numbers".

For the range, though, you can assume that the squared part has to be greater than or equal to 0. That part, (x-3)^2, can't be negative. So, we would have:

(x-3)^2 >= 0

Then, to make it like the function we have, we can multiply each side by 2:

2(x-3)^2 >= 0

Then, add 1 to each side

2(x-3)^2 + 1 >= 1

The left side is y. So:

y >= 1.

And, y represents the range. So, for this function, the range has to be greater than or equal to 1.

Good luck, super. I hope this helps.

Till Then,

Steve

### Hide Replies ▲

Why do we have to assume for the range that the squared bracket must be = or greater than zero?

Thus is graph for the function

`y=2(x-3)^2+1`

It is well defined for all values of x ( you may think of this because there is no radical sign on y).Thus domain of y is all real numbers.

Range : minimum value of y is 1 (if x=3),graph is moving up and up.

so range of y is all real numbers greter or equals to 1.