Question

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

Answer 1

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.