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Mathematics 13 Online
OpenStudy (anonymous):

Use transformations and zeros of the quadratic function f(x) =(x+2)(x-6) to determinate the zeros.

OpenStudy (anonymous):

e) y=-f(x+1) and f) y=f(-x-2)

OpenStudy (rulnick):

(e) y = -f(x+1) = -(x+3)(x-5) has zeros at x=-3 and x=5

OpenStudy (rulnick):

(f) y = f(-x-2) = (-x)(-x-8) has zeros at x=0 and x=-8.

OpenStudy (anonymous):

but how did you get that answer? the steps?

OpenStudy (rulnick):

Well, for example, in (e) replace x with x+1 in the original expression for f(x), then make the whole thing negative.

OpenStudy (anonymous):

whole thing negative? so it's going to be y= -f(x+1) -> y= (-x-1)?

OpenStudy (rulnick):

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?

OpenStudy (anonymous):

ohhh, yes thank you!

OpenStudy (rulnick):

welcome.

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