How do you find the domain and range of #f(x) =x^2+8x+15#?

1 Answer
Feb 20, 2018

Answer:

Domain: #(-oo, oo)#

Range: #[-1, oo)#

Explanation:

Given:

#f(x) = x^2+8x+15#

First note that in common with any polynomial, the (implicit) domain of #f(x)# is the whole of the real numbers #(-oo, oo)#.

We can find the range by completing the square:

#x^2+8x+15 = x^2+8x+16-1 = (x+4)^2-1#

Note that #(x+4)^2 >= 0# for any real value of #x#, taking the value #0# when #x=-4#. So #f(x)# takes its minimum value #-1# when #x=-4#.

Then since #f(x)# is continuous and increases without limit as #x->+-oo#.

So the range is #[-1, oo)#.

Another way of finding the range is to set #y = f(x)# then solve for #x#...

Given:

#y = x^2+8x+15 = (x+4)^2-1#

Add #1# to both ends to get:

#y + 1 = (x+4)^2#

Take the square root of both sides, allowing for both possibilities to get:

#x+4 = +-sqrt(y+1)#

Subtract #4# from both sides to get:

#x = -4+-sqrt(y+1)#

So for any #y >= -1#, there is at least one value of #x# such that #f(x) = y#. In other words, any #y in [-1, oo)# is part of the range, and no other values are.