# How do I use completing the square to describe the graph of f(x)=30-12x-x^2?

Sep 1, 2015

$f \left(x\right) = 30 - 12 x - {x}^{2}$
$= - \left(36 + 12 x + {x}^{2} - 36\right) + 30$
$= - {\left(x + 6\right)}^{2} + 66$

To plot the graph, find $f \left(x\right) = 0$, $f \left(0\right)$ and local maxima/minima.

y-axis intercept:
$f \left(0\right) = 30$

x-axis intercept:
$f \left(x\right) = 0$
${\left(x + 6\right)}^{2} = 66$
$x + 6 = \pm \sqrt{66}$
$x = - 6 \pm \sqrt{66}$

Since this is a quadratic equation there is one stationary point. In this case it is a maximum since the coefficient of highest order $x$ term is negative. This happens when:
${\left(x + 6\right)}^{2} = 0$
So:
$x = - 6$, $f \left(- 6\right) = 66$

graph{30-12x-x^2 [-20, 10, -100, 100]}