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The graph of the parabola defined by the equation y=(x-2)^2+3 is rotated 180 degrees about its vertex, then shifted 3 units to the left, t

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The graph of the parabola defined by the equation y=(x-2)^2 + 3 is rotated 180 degrees about its vertex, then shifted 3 units to the left, then shifted 2 units down. The resulting parabola has zeros at x=a and x=b. What is a+b?

Help me solve this please! Thanks

Jan 4, 2018
edited by Guest  Jan 4, 2018

#1
+1

y  =  (x - 2)^2 + 3

This parabola has its vertex at  (2, 3) . Rotating it  180°  about its vertex has the same effect as flipping it over the x-axis and then shifting it up  6  units. So our new parabola has the equation...

y  =  -[ (x - 2)^2 + 3 ] + 6

Now let's shift it  3  units to the left...

y  =  -[ (x - 2 + 3)^2 + 3 ] + 6

And  2  units down...

y  =  -[ (x - 2 + 3)^2 + 3 ] + 6 - 2          Then simplify.

y  =  -(x +1)^2 + 1

The zeros of this parabola are the  x  values when...

0  =  -(x +1)^2 + 1

(x +1)^2  =  1

x + 1   =   ±√1

x + 1  =  1          or         x + 1  =  -1

x  =  0                or         x  =  -2

And the sum of the zeros   =   0 + -2   =   -2

Here's a graph of the original parabola and the transformed parabola.

Jan 4, 2018
edited by hectictar  Jan 4, 2018
edited by hectictar  Jan 4, 2018
#2
+2

y = (x - 2)^2  + 3

If this is rotated 180° about its vertex, the vertex does not change..........the new function is

y  =  - (x- 2)^2  + 3

Shifting this to the left 3 units left and 2 units down  results in the function

y  =  - (x + 1)^2   +  1

To find the zeroes, we have

-(x + 1)^2  +  1  =  0

-(x +1)^2  =  -1

(x + 1)^2  =  1

x + 1  =  ±1

So

x +  1  =  1 ⇒  x  = 0

x + 1  =  - 1  ⇒  x  = -2

So    a  +  b  =    0   +   -2    =    - 2   