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

Question

Ethan Adkins · Accepted Answer

Domain: #(-\infty , \infty)#
Range: #(-\infty, 3]# We must look for divide by zeroes and negatives under radicals in order to locate the domain. 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]#.

Eva Adamson · Answer

undefined The domain of $y = -(x - 2)^2 + 3$ is all real numbers, $(-∞, ∞)$, because there are no restrictions on the input $x$.

The range of $y = -(x - 2)^2 + 3$ is $y ≤ 3$ because the maximum value of $-(x - 2)^2$ is 0, and when added to 3, it results in a range of values less than or equal to 3.