SOLUTION: Find the equation of the parabola with vertex (4,-4), the parabola opens upwards and is congruent to y=1/2x^2. Can you please help me? Thanks so much in advance:)

Question 703939: Find the equation of the parabola with vertex (4,-4), the parabola opens upwards and is congruent to y=1/2x^2.
Can you please help me? Thanks so much in advance:)
Found 2 solutions by DrBeeee, josmiceli:
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
I don't think you mean congruent because the means they are equal. I think you means same "shape" but shift the vertex from (0,0) to (4.-4).
Given
(1) y = 1/2*x^2 whose vertex is at (0,0).
To shift (1) down by 4 units, subtract 4 from (1) and get
(2) y = 1/2*x^2 - 4
To shift (2) to the right by 4 units subtract (not add) 4 units from x in (2) and get
(3) y = 1/2*(x-4)^2 - 4
Now FOIL (x-4)^2 and (3) becomes
(4) y = 1/2*(x^2 - 8*x +16) -4 or
(5) y = 1/2*x^2 - 4*x +8 -4 or
(6) y = 1/2*x^2 - 4*x + 4
Let's check this to see if the point (4,-4) is satisfied.
Is (-4 = 1/2*4^2 - 4*4 + 4)?
Is (-4 = 8 - 16 + 4)?
Is (-4 = -8 + 4}?
Is (-4 = -4)? Yes
Answer: The equation of the parabola that has its vertex at (4.-4) and opens upward with a compression factor of 1/2 is
y = 1/2*x^2 - 4*x +4.

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The given parabola, , opens upward ( has a minimum ),
and it's vertex is at the origin, (0,0)
Now you just have to translate the vertex to ( 4,-4 )

Here are plots of both equations:

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