## Bounds of the Reciprocal of a Quadratic

The reciprocal of a quadratic, if the quadratic is never zero, is bounded. We can find the bounds by completing the square for the quadratic.
Suppose we want to find the bounds of
$\frac{2}{x^2+4x+9}$
.
Numerator and denominator are both positive, and as
$x \rightarrow \infty 0$
, the denominator tends to infinity also, so the fraction tends to zero. Hence
$0 \lt \frac{2}{x^2+4x+9}$
.
Completing the square for the denominator gives
$\frac{2}{x^2+4x+9}=\frac{2}{(x+2)^2+5}$
.
To maximise this fraction we must minimise the denominator, which is the sum of non negative terms. The denominator is minimised when
$(x+2)^2=0 \rightarrow x=-2$
.
For this value of
$x$
the fraction is equal to
$0 \lt \frac{2}{(-2+2)^2+5}=\frac{2}{0+5}=\frac{2}{5}$
.
Hence
$0 \lt \frac{2}{x^2+4x+9} \lt \frac{2}{5}$
.
The graph of the function is shown below.