How do you find the intercepts, vertex and graph #f(x)=x^2-9#?

1 Answer
bp
Jan 21, 2017

vertex (0,-9)

x intercepts (3,0) and (-3,0)

y intercept (0,-9)

Explanation:

#f(x)=y= x^2-9# represents a vertical parabola opening up. Written in standard form as #(y+9)= (x-0)^2#, shows that its vertex is at (0,-9).

For intercepts, let x=0, which gives intercept y=-9. For x intercept put y=0, which gives #x^2 =9#. Thus there would be two x -intercepts (3,0) and (-3,0). The graph of parabola would be like this :

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