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

1 Answer
George C.
Jun 7, 2015

(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.

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