Math CentralQuandaries & Queries


Question from linda, a parent:

I have been trying to solve this problem but I'm unable to figure it out.  How do I find the domain and range of y=-(x+1)^2-3?  Please explain...thanks!

Hi Linda.

The domain of this function is the set of all possible values of x which you can put in the equation and get a value for y.  It should be clear to you that you could conceivably put any value of x in and you'll get something out.  So the domain is the set of all real numbers.

The range is all the possible values of y.  That's more involved. Let's first move things around a little.

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

Now let's stop here.  The value on the right hand side is a perfect square, so it can never be negative. That is the same as saying it is greater than or equal to zero. Since (x+1)2 ≥ 0, the left hand side is also.  Solve for y to complete:

-3 - y ≥ 0
-3 ≥ y
y ≤ -3

Hence y must be less than or equal to minus 3. I want to show that every such number is in the range of the function. To do so I need to take a number y which is less than or equal to -3 and find a number x so that -3 - y = (x+1)2.

Ley y be a number that satisfies y ≤ -3 then -3 - y ≥ 0 and hence I can take the square root of -3 - y since it's not negative. Let x = -1 + √(-3 - y) and then x + 1 = √(-3 - y) and hence (x+1)2 = -3 - y. Thus the range is all numbers equal to or less than -3.

There are several ways of expressing the domain and range.  Here's how I would write it:

domain: (-∞, +∞), range: (-∞, -3].  

By convention, round parentheses means "not including the end point" and square brackets mean "including the end point". Infinities are always round parentheses.

Hope this helps,
Stephen La Rocque and Penny Nom.

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