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Bunuel
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Bunuel
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siriusly
Hi Bunuel, could you provide the solution? Thank you!
I have no paper here, but just plugin this in the answer:
Y=0
X=-3

The one that gets you 0=0 is the right one
Therefore, e

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\(y = \frac{x^2}{3}\), and the shape of this graph is a parabola. As per the question, "translated three units horizontally and to the left". Only the \(x\) coordinates should be translated horizontally and to the left by \(3\) units, for e.g. if the coordinates are \((x,y)\), then \((x-3,y)\).
In Parabola, we need to shift the graph to the left by 3 units as shown in the attached screenshot. The Parabola in "Green" i.e. \(y = (x+3)^2\)shows the shift by 3 units to the left for the equation \(y = \frac{x^2}{3}\).
Ans E

Note: Please observe the graph for better understanding.
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parabola 2.jpg
parabola 2.jpg [ 129.88 KiB | Viewed 2095 times ]

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Bunuel
In the xy-plane, which of the following is an equation whose graph is the graph of y = x^2/3 translated three units horizontally and to the left?


(A) y = x^2

(B) y = x^2/3 + 3

(C) y = x^2/3 - 3

(D) y = (x - 3)^2

(E) y = (x + 3)^2

Solution:

When the graph of y = f(x) is translated k (where k > 0) units horizontally to the left, the new graph has the equation y = f(x + k). Here, we have y = f(x) = x^2/3 and k = 3; therefore, the new graph has the equation y = f(x + 3) = (x + 3)^2/3.

(Note that no answer choice is correct, but it is fairly obvious that choice E is the intended correct answer. It shows the horizontal translation correctly, but it does not contain the coefficient a = ⅓ that was present in the original function. We see that choices B and C display a vertical shift and choices D and E display a horizontal shift. Choice D is a shift to the right, and choice E displays the desired shift to the left.)

Answer: E
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