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anzhelika [568]
2 years ago
10

The function f(x) = x2 6x 3 is transformed such that g(x) = f(x − 2). find the vertex of g(x). (−1, −8) (−3, −4) (−1, −6) (−5, −

6)
Mathematics
1 answer:
timama [110]2 years ago
3 0

Answer: (-1,-6)

Step-by-step explanation:

x^2+6x+3

g(x) = f(x-2)

replace all value of x in the f(x) function to (x-2)

(x-2)^2 + 6(x-2) + 34

(x-2)(x-2) + 6(x-2) + 34

expand through FOIL and distribution

x^2-4x+4+6x-12+3

combine like terms

x^2+2x-5

-b/2a = x value of vertex

-2/2(1) = x value

x value = -1

plug in -1 for all values of x in the g(x) equation to solve for y value

(-1)^2+2(-1)-5

1-2-5 = -6

y = -6

x = -1

(x,y) = (-1,-6)

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Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

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Step-by-step explanation:

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\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

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And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

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And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

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E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

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