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DochEvi [55]
2 years ago
12

Given the function f(x) = 3x - 1, explain how to find the average rate of change between x = 1 and x = 4.

Mathematics
2 answers:
Mice21 [21]2 years ago
6 0

Step-by-step explanation:

f(1) = 3×1 - 1 = 2

f(4) = 3×4 - 1 = 12-1 = 11

so, the functional value changes 11-2=9 units on an x interval of 4-1=3 units length.

the average change rate is the total change across the x interval relative to the interval length.

that is

9/3 = 3

which is the slope (= the factor of x) in the line equation.

for a line its change rate for any point is the same constant. and that is therefore automatically also the average change rate across an interval of x values.

if the change rate would be different for different parts of the function, it would not be a straight line.

Tems11 [23]2 years ago
6 0

Answer:

3

Step-by-step explanation:

The average rate of f(x) in the closed interval [ a, b ] is

\frac{f(b)-f(a)}{b-a}

Here [ a, b ] = [ 1, 4 ] , then

f(b) = f(4) = 3(4) - 1 = 12 - 1 = 11

f(a) = f(1) = 3(1) - 1 = 3 - 1 = 2

average rate of change = \frac{11-2}{4-1} = \frac{9}{3} = 3

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Answer:

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

  (x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

  also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

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- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵  (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

   the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

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∴ The standard form of the equation of the hyperbola is

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