What is the range of #f(x)=–5-2(x+3)^2#?

1 Answer
Dec 31, 2016

Answer:

The range is #{y|y<="-5"}#.

Explanation:

The range of a function is simply all the possible outputs that function can give.

Mathematically speaking, a number #y# is in the range of a function #f# when we can find an #x# such that #f(x)=y#.

There are a couple of ways to find the range of a function. For #f(x)="-5"-2(x+3)^2#, the easiest way is to see that #f# is a certain type of function—a parabolic function. As such, there is no limitation on the #x#-values we can input (i.e. the domain is #RR#), but there is an easy-to-see limitation on the output (or #y#) values #f(x)# can give.

Take a look at that squared bit—the #(x+3)^2#. We know this can't be less than 0, because the square of a number is always positive (or 0, if we're squaring 0).

So #(x+3)^2# is no less than 0. Meaning, #2(x+3)^2# is also no less than 0. Then #-2(x+3)^2# is no more than 0, and thus #"-"5-2(x+3)^2# is no more than -5.

In math:

#color(white)(f(x)="-5"-2)(x+3)^2>=0#
#color(white)(f(x)="-5"-)2(x+3)^2>=0#
#color(white)(f(x)="-5")-2(x+3)^2<=0" "#(note the change to #<=#)
#f(x)="-5"-2(x+3)^2<="-5"#

The end result: #f(x)<="-5"#.
And so our range is "all numbers #y# such that #y# is at most -5", or in math:

#{y|y<="-5"}#.

Bonus:

The shortcut to finding the range of a parabolic equation #f(x)=a(x-h)^2+k# is to use this:

Range of #f(x)# is #{({y|y>=k}" when "a>0),({y|y<=k}" when "a<0):}#

Just choose the correct option depending on the value of #a#. (In this example, #a# was less than 0, so we would choose #{y|y<=k}#, and plug in our #k#-value of -5.)