How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #y = -x^2 + 6x -9#?

1 Answer
Bill K.
Oct 15, 2015

This one is easy to factor to find the vertex to be the point #(x,y)=(3,0)# so that #x=3# is the axis of symmetry. The #y#-intercept is #-9#, the #x#-intercept is #x=3#, the domain is all of #RR#, and the range is #(-infty,0]#.

Explanation:

You can factor #y=f(x)=-x^2+6x-9# as the opposite of a perfect square (there's no need to "complete the square" here):

#y=-x^2+6x-9=-(x^2-6x+9)=-(x-3)^2#.

This means the parabola opens downward with a vertex (high point in this case) at #(x,y)=(3,0)# (all other values of the function are negative). This also gives #x=3# as the axis of symmetry and the (one) #x#-intercept.

For the #y#-intercept, compute #f(0)=-0^2+6*0-9=-9#.

The domain is the entire real number system #RR# because you can plug in anything you want for #x#. The range is the set of non-positive numbers #(-infty,0]# because that's the set of possible outputs.

The graph is shown below. Make sure you think about all these answers in relation to the graph.

graph{-x^2+6x-9 [-40, 40, -20, 20]}

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