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3 years ago

What is 2x-4. I really don’t understand

Mathematics

1 answer:

ddd [48]3 years ago

3 0

Answer:

2 * -4 = -8

Step-by-step explanation:

2 * -4 = -8

It is just like multiplying two numbers like... 2 * 4 = 8

You must need to know that:

+a * -b = -c

-a * -b = +c

+a * +b = +c

In your question;

+a * -b = -c

a = 2

b = -4

c = -8

therefore:

2 * -4 = -8

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Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

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Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

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since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

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$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

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Answer:

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Answer and Step-by-step explanation:

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F’(x) = [ x2 + (14 – x)2]’ = 0

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