Question 135308: If the graph of y=x^2 is transformed to y=3(2)^(-1/4x-1)+1, what are the coordinates of the new focal point? I think (-4,4)????
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If the graph of y=x^2 is transformed to y=3(2)^(-1/4x-1)+1, what are the coordinates of the new focal point? I think (-4,4)????
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Rearranging the transformed form you get:
(1/3)(y-1)= 2^[(-1/4)(x + 4)]
VS = 3, VT = 1 ; HS=-4,HT=-4
Mapping: (x,y) goes to (HSx+Ht,VSy+VT)
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The focal point for y = x^2 is (0,1/4)
Your mapping form is (x,y) goes to (-4x-4,3y+1)
So (0,1/4) goes to (-4,7/4)
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Cheers,
Stan H.
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