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f(x)=(x+5)^2(x-3)^2 for intercepts I got: x (-5,0),(3,0) y (0,225)
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Hmm... May I know what your question is?
did i do this correct
Yes. For x-int, put f(x) = 0 0 = (x+5)^2(x-3)^2 (x+5)^2=0 or (x-3)^2=0 x = -5 or x = 3 So, x-int are at ( -5, 0) and (3, 0) For y-int, y-int = f(0) = (0+5)^2(0-3)^2 = 5^2 ( -3)^2 = 25 (9) = 225 So, y-int is at (0, 225)
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