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Butoxors [25]
2 years ago
10

The function f(x) = x2 has been translated 9 units up and 4 units to the right to form the function g(x). Which represents g(x)?

Mathematics
2 answers:
tester [92]2 years ago
8 0
Hello the answer is 

<span>c)g(x) = (x − 4)2 + 9 
</span>
Alenkinab [10]2 years ago
7 0

For this case, the parent function is given by:

f (x) = x ^ 2

We apply the following transformations:


Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units upwards:

For k = 9 we have:

h (x) = x ^ 2 + 9


Horizontal translations:

Suppose that h> 0

To graph y = f (x-h), move the graph of h units to the right

For h = 4 we have:

g (x) = (x-4) ^ 2 + 9

Answer:

The function g (x) is given by:

g (x) = (x-4) ^ 2 + 9

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VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

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b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

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y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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kumpel [21]

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The coordinate (-3,-9) is all negative, therefore it lies in quadrant III.

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2 years ago
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