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Use transformations and zeros of the quadratic function f(x) =(x+2)(x-6) to determinate the zeros.
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e) y=-f(x+1) and f) y=f(-x-2)
(e) y = -f(x+1) = -(x+3)(x-5) has zeros at x=-3 and x=5
(f) y = f(-x-2) = (-x)(-x-8) has zeros at x=0 and x=-8.
but how did you get that answer? the steps?
Well, for example, in (e) replace x with x+1 in the original expression for f(x), then make the whole thing negative.
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whole thing negative? so it's going to be y= -f(x+1) -> y= (-x-1)?
No. When you replace x with x+1 you get (x+1+2)(x+1-6) = (x+3)(x-5). Making this negative, -(x+3)(x-5). See this in my first post?
ohhh, yes thank you!
welcome.
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