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The graph of the parabola defined by the equation y=-(x+1)^2+1 is shifted 1 unit to the right, then shifted 5 units down. The resulting parabola has zeros at x=a and x=b, where b≥a. What is b-a?

 Jan 14, 2021
 #1
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Lead:

When y = -(x+1)^2 + 1 is shifted 1 unit to the right, the vertex moves 1 un right as well. So in vertex form, the equation becomes y = -(x)^2 + 1. Then, shifting 5 units down with similar logic gets us y = -(x)^2 - 4.

Now we find the zeros of this equation. Expanding we get y = -x^2 - 4, so dividing by -1 gets us x^2 + 4 = 0. Can you go from here?

 

You are very welcome!

:P

 Jan 14, 2021
 #2
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Hmmmm..... shifting this parabola right one and down five results in no REAL zeroes....only imaginary zeroes  ± 2i

 

 Jan 14, 2021

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