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Brilliant_brown [7]
3 years ago
9

Jacques needs to convert the function f(x) = x2 − 6x + 14 to vertex form, f(x) = (x − h)2 + k, in order to find the minimum.

Mathematics
2 answers:
Olenka [21]3 years ago
6 0

Answer:

C. 5

Step-by-step explanation:

Given function is,

f(x) = x^2 - 6x + 14

Here, the coefficient of x = 6,

Thus, We need to add and subtract the square of half of 6 in the given equation for getting the vertex form,

By adding and subtracting 9,

f(x) = x^2 - 6x + 9 + 14 - 9

f(x) = (x-3)^2 + 5   ( Because, a² - 2ab + b² = (a-b)²  )

Which is the required vertex form he will get,

By comparing it with f(x) = (x+h)^2+k

We get, k = 5,

⇒ Option C is correct.

Anarel [89]3 years ago
5 0

k = 5

the equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

To obtain this form use the method of completing the square

Since the coefficient of the x² term is 1 then

add/ subtract (half the coefficient of the x-term )² to x² - 6x

f(x) = x² + 2(- 3)x + 9 - 9 + 14 = (x - 3)² + 5 → k = 5


   

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taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

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Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

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Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

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* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

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∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

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