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

1 Answer
Sep 1, 2015

# 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]}