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madam [21]
3 years ago
8

What are the equations of the asymptotes of the rational function? Please explain. f(x)=

c%7B5%7D%7Bx%2B7%7D%2B6" id="TexFormula1" title="\frac{5}{x+7}+6" alt="\frac{5}{x+7}+6" align="absmiddle" class="latex-formula">
A. x=-7 and y=6
B. x=5 and y=6
C. x=7 and y=-6
Mathematics
2 answers:
Nady [450]3 years ago
4 0

\bf f(x)=\cfrac{5}{x+7}+6\implies f(x)=\cfrac{5+6(x+7)}{x+7}\implies f(x)=\cfrac{6x+47}{1x+7}


to get the vertical asymptotes, we simply zero out the denominator and see what's the equation so let's do so

x + 7 = 0............. x = -7....... that's the vertical asymptote.


to get the horizontal asymptotes, there are 3 cases

if the denominator's degree is higher, or lower or the exactly the same as the degree of the numerator.

well, notice the rational, the degree of the numerator is just 1, 6x¹, and the degree of the denominator is also 1, 1x¹.

when that occurs, the horizontal asymptote occurrs at the fraction of their leading term coefficients.


\bf f(x)=\cfrac{6x+47}{1x+7}~\hspace{10em}\stackrel{\textit{horizontal asymptote}}{y=\cfrac{6}{1}}\implies y=6 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{V~asymptote}{x=-7}\qquad \stackrel{H~asymptote}{y=6}~\hfill

tangare [24]3 years ago
3 0

Answer:

vertical asymptote at x = 7

horizontal asymptote at y = 6

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.


solve:  

x − 7 = 0 ⇒ x = 7  is the asymptote

Horizontal asymptotes occur as

lim ,f(x)→ c(a constant)

x → ± ∞

divide terms on numerator/denominator by x

f ( x ) = 5/ x +6= 5 /x + 6

x/ x − 7 /x      1 − 7/ x

as   x ± ∞ , f ( x ) → 0 +6

                1 − 0

⇒ y = 6  is the asymptote

graph{((5)/(x-7))+6 [-20, 20, -10, 10]}

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3 years ago
Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
Levart [38]

A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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1 year ago
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2 years ago
You walk into a store that is closing. the pants normally cost $60.00 but it is discounted 75% they are also taking additional 2
eimsori [14]

Answer:

The final cost of the pants is $ 12.

Step-by-step explanation:

Given that at a store that is closing the pants normally cost $ 60.00 but it is discounted 75%, and they are also taking an additional 20% off, to determine what is the final price for the pants, the following calculation must be performed:

100 - 20 = 80

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2 years ago
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d. converges, -25

Step-by-step explanation:

An infinite geometric series converges if the absolute value of the common ratio is less than 1.

Here, the common ratio is 4/5:

| 4/5 | = 4/5 < 1

So the series converges.  The sum of an infinite geometric series is:

S = a₁ / (1 − r)

where a₁ is the first term and r is the common ratio.

Here, a₁ = -5 and r = 4/5:

S = -5 / (1 − 4/5)

S = -25

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