Anyone mind helping me understand this?

Given the functions a(x) = 4x2 + 2x − 3 and b(x) = x − 1, identify the oblique asymptote of the function a of x over the function b of x

Question

Anyone mind helping me understand this? 

Given the functions a(x) = 4x2 + 2x − 3 and b(x) = x − 1, identify the oblique asymptote of the function a of x over the function b of x

michele_laino · Answer

we have to write the oblique asymptotes of this function:
$$f\left( x \right) = \frac{{a\left( x \right)}}{{b\left( x \right)}} = \frac{{4{x^2} + 2x - 3}}{{x - 1}}$$

welshfella · Answer

Do the long division
- what do you get?

michele_laino · Answer

first step
we have to evaluate this limit:
$$\mathop {\lim }\limits_{x \to  \pm \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to  \pm \infty } \frac{{4{x^2} + 2x - 3}}{{x\left( {x - 1} \right)}} = ...?$$

zoiedaniella · Answer

that will be 4x+6+3/x-1 simplified correct?

michele_laino · Answer

hint:
divide numerator an denominator by $x$

zoiedaniella · Answer

mmkay, lemme do this on paper real quick, brb.

michele_laino · Answer

starting from your result, of course

zoiedaniella · Answer

okay, do you mind writing out what I would divide? I just really confused myself for a min there.

michele_laino · Answer

mean this:
starting from your expression:
(4x+2-(3/x))/(1-(1/x))
please divide again numerator and denominator by x

michele_laino · Answer

sorry, 
please start from this function:
$$\frac{{4{x^2} + 2x - 3}}{{{x^2} - x}}$$
and divide numerator and enominator by $x^2$

michele_laino · Answer

denominator*

zoiedaniella · Answer

that would just cancel out the x^2's correct?

zoiedaniella · Answer

so it would just be 4+2x-3/-x ?

zoiedaniella · Answer

and now do long division?

michele_laino · Answer

I got this:
for numerator
$$4{x^2} + 2x - 3 \to 4 + \frac{2}{x} - \frac{3}{{{x^2}}}$$

michele_laino · Answer

now, do the same with the denominator please

zoiedaniella · Answer

$$x^2/x^2 -x/x^2 $$ ?? i'm sorry i'm very confused haha

michele_laino · Answer

correct! SO, after a simplification we get:
$$\frac{{{x^2} - x}}{{{x^2}}} = 1 - \frac{1}{x}$$

michele_laino · Answer

oops.. So*...

zoiedaniella · Answer

ohhhh okay

michele_laino · Answer

now we have to evaluate this limit:
$$\mathop {\lim }\limits_{x \to  \pm \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to  \pm \infty } \frac{{4{x^2} + 2x - 3}}{{x\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to  \pm \infty } \frac{{4 + \frac{2}{x} - \frac{3}{{{x^2}}}}}{{1 - \frac{1}{x}}} = ...?$$

michele_laino · Answer

hint:
we have this:
as $x$ goes to infinity, the quantity $1/x$ goes to zero

michele_laino · Answer

and also the quantity $1/x^2$ goes to zero

zoiedaniella · Answer

doing it on paper, one min. 
So the answer would be $$4+(6x-3)/x^2-x$$

michele_laino · Answer

the answer is:
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