How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2-9#?

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Answer 1 · 2017-07-05T11:47:53

Vertex is at # (0,-9)# , y-intercept: #y= -9 or (0, -9)# ,
axis of symmetry is #x=0# or y-axis.

#f(x) = x^2 -9 = a(x-0)^2 -9 #. Comparing with standard vertex form

of equation #f(x)=a(x-h)^2+k ; (h,k)# being vertex,

we find here #h=0,k= -9#. So vertex is at #(h,k) or (0,-9)#.

putting #x=0# in the equation we find y-intercept as

#y= 0^2-9 = -9 or (0, -9)#

Axis of symmetry is #x=0# or y-axis.

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