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Serggg [28]
11 months ago
9

Instructions: For the function given, determine the direction and amount of vertical shift from the function =(2).

Mathematics
2 answers:
posledela11 months ago
5 0

SOLUTION

From the translation given, the amount of vertical shift is 3 units down

This was derived from -3

Hence the answer is 3 units down

Lemur [1.5K]11 months ago
4 0
The function is translated 1 unit to the left, 3 units down, and is stretched by a factor of 6.

Translated functions use their principle (original) function and add lateral and vertical changes to the function. The principle function is y=(2)^x. Therefore, adding one to that x-value would shift it to the left by 1 unit for every x-value. NOTE: +1 is actually to the right, but transformations reverse that, so it is actually to the left towards the negative domains. The entire function is shifted down 3 units; the “-3” as the last term is subtracting the entire function, so the y-coordinates will all shift down 3 units. And the “a” value of the function is 6. When a>0, the function stretches. So, the function will stretch by a factor of 6, since it is multiplying the entire range of the function.
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Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
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