What is the domain and range of #y+2 = (x-3)^2#?

1 Answer
Bassel T
Jun 14, 2018

Domain: #x inRR#
Range: #y in [-2,oo)#

Explanation:

The function you provided is almost in vertex form of a quadratic function, which helps greatly when answering your question. Vertex form in a quadratic is when the function is written in the following form:

#y=a(x-h)^2+k#

To write your function in vertex form, I'll simply solve for #y# by subtracting 2 from both sides:

#y=(x-3)^2-2#

The two parameters you want in this are #a# and #k#, since those will actually tell you the range. Since any value of #x# can be used in this function, the domain is:

#x inRR#

Now we need the range. As stated before, it comes from the values of #a# and #k#. If #a# is negative, the range goes to#-oo#. If #a# is positive, the range goes to #oo#. In this case, #a# is positive, so we know the range goes to #oo#. The lowest value will be the #k# value, which in this case is -2. Hence, the range of your function is:

#y in [-2,oo)#

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