Five sixths divided by four = one over blank of five sixths

Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4

djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

$\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty$

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

$\lim_{x \to \infty}\frac{x^3}{(x-2)^4}$

if $\lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b$ then y = b is horizontal asymptote

Then:

$\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0$

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=7y%20%5Ctimes%204x%20-%205x%20%7B%3F%7D%5E%7B2%7D%20" id="TexFormula1" title="7y \times 4x - 5

lapo4ka [179]

The answer is i don’t know or I don’t care and what are we trying to find

5 0

3 years ago

A receptionist named Eve spends 5minutes routing each incoming phone call. In all,how many phone calls does Eve have to route to

Sati [7]

7, because 35 divided by 5=7

5 0

3 years ago

Read 2 more answers

Ms Angelina made 2 pans of lasagna and cut each pan into twelfths her family ate 1 1/2 pans of lasagna for dinner how many pans

Furkat [3]

Answer:

1/2 of the lasagna was left or 6/12

Step-by-step explanation:

Her family ate 1 whole pan of the first lasagna and 1/2 of the second pan. If they ate 1/2 from the second pan then there will be 1/2 left or in this case 6/12 because she cut it in 12 pieces.

Hope this helped :)

8 0

3 years ago

Match the numerical expressions to their simplest forms.

Aloiza [94]

Answer:

$(a^6b^1^2)^\frac{1}{3}$ = $a^2b^4$

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$ = $a^3b^2$

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$ = $a^2b$

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$ = $ab^3$

Step-by-step explanation:

Simplify each of the expressions:

1

$(a^6b^1^2)^\frac{1}{3}$

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

$(a^6b^1^2)^\frac{1}{3}$

$a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}$

Simplify,

$a^2b^4$

2

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

$\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}$

Rewrite as square roots:

$\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}$

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

$\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}$

Simplify,

$\sqrt{a^5b^3}*\sqrt{ab}$

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

$\sqrt{a^5b^3*ab}$

Simplify,

$\sqrt{a^6b^4}$

Since this operation is taking the square root, divide the exponents in half to do this operation:

$a^3b^2$

3

$(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}$

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

$(\frac{a^8}{b^-^4})^\frac{1}{4}$

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

$(a^8b^4)^\frac{1}{4}$

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

$\sqrt[4]{a^8b^4}$

SImplify, divide the exponents by (4) to simulate taking the quartic root,

$a^2b$

4

$(\frac{a^3}{ab^-^6})^\frac{1}{2}$

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

$(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}$

$(a^2b^6)^\frac{1}{2}$

Rewrite to the half-power as a square root,

$\sqrt{a^2b^6}$

Simplify, divide all of the exponents by (2),

$ab^3$

7 0

3 years ago