Question

What is the range of the quadratic function `f(x) = 5x^2 + 20x + 4`?

Answer 1

`(x+2)^2 = x^2+4x+4`

`5(x+2)^2 = 5x^2+20x+20`

So

`f(x) = 5x^2+20x+4`

`= 5x^2+20x+20-16`

`= 5(x+2)^2-16`

The minimum value of `f(x)` will occur when `x=-2`

`f(-2) = 0-16 = -16`

Hence the range of `f(x)` is `[-16, oo)`

More explicitly, let `y = f(x)`, then:

`y = 5(x+2)^2 - 16`

Add `16` to both sides to get:

`y + 16 = 5(x+2)^2`

Divide both sides by `5` to get:

`(x+2)^2 = (y+16)/5`

Then

`x+2 = +-sqrt((y+16)/5)`

Subtract `2` from both sides to get:

`x = -2+-sqrt((y+16)/5)`

The square root will only be defined when `y >= -16`, but for any value of `y in [-16, oo)`, this formula gives us one or two values of `x` such that `f(x) = y`.