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

Jun 7, 2015

${\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4$

$5 {\left(x + 2\right)}^{2} = 5 {x}^{2} + 20 x + 20$

So

$f \left(x\right) = 5 {x}^{2} + 20 x + 4$

$= 5 {x}^{2} + 20 x + 20 - 16$

$= 5 {\left(x + 2\right)}^{2} - 16$

The minimum value of $f \left(x\right)$ will occur when $x = - 2$

$f \left(- 2\right) = 0 - 16 = - 16$

Hence the range of $f \left(x\right)$ is $\left[- 16 , \infty\right)$

More explicitly, let $y = f \left(x\right)$, then:

$y = 5 {\left(x + 2\right)}^{2} - 16$

Add $16$ to both sides to get:

$y + 16 = 5 {\left(x + 2\right)}^{2}$

Divide both sides by $5$ to get:

${\left(x + 2\right)}^{2} = \frac{y + 16}{5}$

Then

$x + 2 = \pm \sqrt{\frac{y + 16}{5}}$

Subtract $2$ from both sides to get:

$x = - 2 \pm \sqrt{\frac{y + 16}{5}}$

The square root will only be defined when $y \ge - 16$, but for any value of $y \in \left[- 16 , \infty\right)$, this formula gives us one or two values of $x$ such that $f \left(x\right) = y$.