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3 years ago

You graph the function f(x)=-|×|-12

Mathematics

2 answers:

Firlakuza [10]3 years ago

7 0

Answer:

See attachment

Step-by-step explanation:

We want to graph $f(x)=|x|-12$ on the interval -10 to 10.

Let $g(x)=|x|$ be the parent absolute value function.

We can easily graph $f(x)=|x|-12$ , if we use translation.

When the parent function is shifted downwards by 12 units, we obtain the graph of $f(x)=|x|-12$ .

The parent function is a v-shaped graph with vertex at the origin.

We shift the parent function down so that its vertex is now at (0,-12) to get the graph of $f(x)=|x|-12$ .

See attachment for the graph of $f(x)=|x|-12$ on the specified interval.

12345 [234]3 years ago

6 0

Answer:

Which of the following did you include in your response?

No, you will not see the graph.

When x is 0, y is –12, which is outside the viewing window.

Because the function has been reflected, it opens down.

From x = –10 to x =10, the y-values range from –22 to –12.

Step-by-step explanation:

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By closure property of multiplication and addition of integers,

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From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

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By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

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