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dem82 [27]
2 years ago
10

If rectangle ABCD, defined by the coordinates A(0, 0), B(0, -4), C(-2, -4), D(-2, 0), is dilated by a scale factor of 2, with re

sulting vertex A' at (2, 4), what is the center of the dilation? A) (0, 0) Eliminate B) (2, 4) C) (0, -4) D) (-2, -4)
Mathematics
1 answer:
notsponge [240]2 years ago
6 0

Answer:

D) (-2, -4).

Step-by-step explanation:

Let (x, y) be the center of dilation. Imagine that (x, y) is the center of origin of a new Cartesian Plane. What would the coordinates of point A and A' on the new plane?

  • A: (0 - x, 0 - y) ⇒ (-x, -y).
  • A': (2 - x, 4 - y).

Dilating (-x, -y) about the "origin" of the new plane by a factor of two will give the point (-{\bf 2}x, -{\bf 2}y), also on the new plane.

For coordinates of A' on the original plane,

(-{2}x, -{2}y) = (2 - x, 4 - y).

As a result,

\left \{ \begin{array}{l}-2x = 2 - x\\-2 y = 4 - y\end{array}.

\left \{ \begin{array}{l}-x = 2\\- y = 4\end{array}.

\left \{ \begin{array}{l}x = -2\\y = -4\end{array}.

In other words, {\bf (-2, -4)} is the center of dilation.

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Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati
amid [387]
In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
f(x) = a(x - x_1)(x-x_2)(x-x_3)

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form (x_n , 0)

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say x_1 = 2 , x_2 = 3 , x_3 = 5

So required equation is
f(x) = a (x-2)(x-3)(x-5)
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Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
-5= a (0-0+0 - 30)
-5 = -30a
a =  \frac{-5}{-30} =  \frac{1}{6}
So required equation of cubic polynomial is
f(x) =  \frac{1}{6}(x^3 -10x^2+31x-30)

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.
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