How do you find the domain and range of #y=-(x-2)^2 +3 #?

1 Answer
Apr 27, 2017

Answer:

Domain: #(-\infty , \infty)#
Range: #(-\infty, 3]#

Explanation:

To find the domain we have to check for divide by zeroes and negatives under radicals.

Since there are no fractions, divide by zero is impossible. Since we aren't any radicals (#\sqrt{x}#), it is impossible for that to happen. So we can conclude that we can put any value of x into the function and get an answer.

To get the range, without using calculus (much easier), we have to think of what value of #x# will give use the greatest value of #y#. Since the value of #-(x-2)^2# will always be negative, then 0 will be be the largest number we can get from any value of x. If we solve for this we get the following:
#x-2=0#
#x=2#

Plugging in #x=2# we get
#y(2)=-(2-2)^2+3#
#=-(0)^2+3#
#=0+3#
#=3#

The smallest value of y will then be #-\infty# because the larger #x# gets, the larger #-(x-2)^2# will be, but it will always be negative. Since 3 doesn't really have much of an impact on really big numbers, the range is #(-\infty, 3]#.