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

1 Answer
George C.
Feb 20, 2018

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.

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