MHB Solving the Slant Asymptote of $$\frac{{x}^{3}-5{x}^{2}+4x}{-4{x}^{2}+36}$$

[karush]

$$\frac{{x}^{3}-5{x}^{2}+4x}{-4{x}^{2}+36 }$$

Has a slant asymtope of $mx+b$ of which I got $-\frac{1}{4}x+b$
I couldn't get the b

 

[MarkFL]

What do you get if you perform polynomial long division? The quotient will be the asymptote, as the remainder will tend to zero for values of $x$ having great magnitude.

 

[Greg]

Alternatively,

$$y=mx+b$$

$$b=y-mx=\lim_{x\to\infty}\dfrac{x^3-5x^2+4x}{-4x^2+36}-\lim_{x\to\infty}-\dfrac x4$$

Simplify and evaluate the limit to find $b$. For other functions it may be necessary to evaluate $x\to-\infty$.

 

[karush]

I divided but got $-5{x}^{2 }-13x$ for a remainder?

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Alternatively,

$$y=mx+b$$

$$b=y-mx=\lim_{x\to\infty}\dfrac{x^3-5x^2+4x}{-4x^2+36}-\lim_{x\to\infty}-\dfrac x4$$

Simplify and evaluate the limit to find $b$. For other functions it may be necessary to evaluate $x\to-\infty$.

This is an exercise before limits are introduced

 

[karush]

I still don't know what $b$ is?

 

[MarkFL]

Let's look at the long division:

$$\begin{array}{r}-\tfrac{1}{4}x+\tfrac{5}{4}\\-4x^2+0x+36\enclose{longdiv}{x^3-5x^2+4x+0} \\ -\underline{\left(x^3+0x^2-9x\right)} \hspace{22px} \\ -5x^2+13x+0 \\ -\underline{\left(-5x^2+0x+45\right)} \hspace{-12px} \\ 13x-45 \hspace{-5px} \end{array}$$

Thus, we may write:

$$\frac{x^3-5x^2+4x}{-4x^2+36}=-\frac{1}{4}x+\frac{5}{4}+\frac{13x-45}{-4x^2+36}$$

Now, notice the remainder (linear) over the divisor (quadratic) will tend to zero for values of $x$ having great magnitude, thus the oblique asymptote is the line:

$$y=-\frac{1}{4}x+\frac{5}{4}$$

 

[suluclac]

Mark is correct; $$-\frac{1}{4}x+\frac{5}{4}$$.
When I do the long division, I usually like to verify my answer as such:
$$(-4x^2+36)(-\frac{1}{4}x+\frac{5}{4})+(13x-45)=x^3-5x^2+4x$$

 

[karush]

i was curious about the latex for long division