Question

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

Answer 1

`f(x) = 30 - 12x - x^2`
`= -(36 + 12x + x^2 - 36) + 30`
`= -(x + 6)^2 + 66`

To plot the graph, find `f(x) = 0`, `f(0)` and local maxima/minima.

y-axis intercept:
`f(0) = 30`

x-axis intercept:
`f(x) = 0`
`(x + 6)^2 = 66`
`x + 6 = +- sqrt(66)`
`x = -6 +- 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:
`(x + 6)^2 = 0`
So:
`x = -6`, `f(-6) = 66`

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