To find the tangent line at the given point, we need to first take the derivative of the given function. 

To find the derivative we need to use product rule. Product rule states that we take the derivative of the first function and multiply it by the derivative of the second function and then add that with the derivative of the second function multiplied by the given first function. To find the derivative of each separate function we need to use power rule. 

\(\displaystyle f(x)g'(x)+g(x)f'(x)\)

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent

\(\displaystyle n(x^{n-1})\)

\(\displaystyle f(x)= x^2 + 2x ; f'(x)= 2x + 2\)

\(\displaystyle g(x)= -x^3 +4 ; g'(x) = -3x^2\)

Use power rule and we get : 

\(\displaystyle (x^2+2x)(-3x^2)+(-x^3+4)(2x+2)\)

From here, to find the slope at the given point we plug in "2" for x.

This comes out to equal 

To solve this, find where the function cannot exist. Here, the function cannot exist if the denominator is zero. This happens at x=2 and x=-2. Graph the function on a graphing calculator or by hand to see that the function never crosses these vertical lines. It only gets infinitely close. This is characteristic of vertical asymptotes. 

\(\displaystyle f(x) = \frac{1-2x^3}{x^3-4x+2x}\)

1) To find the horizontal asymptotes, find the limit of the function as \(\displaystyle x->\infty\), 

 \(\displaystyle \lim_{x->\infty}\frac{1-2x^3}{x^3-4x^2+4x}=\lim_{x->\infty}\frac{\frac{1}{x^3}-2}{1-\frac{4}{x}+\frac{4}{x^2}} = -2\)

 Therefore, the function \(\displaystyle f(x)\) has a horizontal asymptote \(\displaystyle y = -2\)

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2) Vertical asympototes will occur at points where the function blows up, \(\displaystyle y->\infty\). For rational functions this behavior occurs when the denominator approaches zero. 

Factor the denominator and set to zero, 

\(\displaystyle x(x^2-4x+4)=x(x-2)^2 = 0\)

\(\displaystyle x = \left \{ 0, 2\right \}\)

So the graph of \(\displaystyle f(x)\) has two vertical asymptotes, one at \(\displaystyle x = 0\) and the other at \(\displaystyle x = 2\).  They have been drawn into the graph of  \(\displaystyle f(x)\) below. The blue curves represent \(\displaystyle f(x)\). 

For this infinity limit, we need to consider the leading terms of both the numerator and the denominator.  In our problem, the leading term of the numerator is larger than the leading term of the denominator.  Therefore, it will be growing at a faster rate. 

\(\displaystyle \lim_{x\rightarrow -\infty}\frac{4x^5-3x^2+2}{2x^3-x^2+1}=\frac{4x^5}{2x^3}=2x^2.\)

Now, simply input the limit value, and interpret the results.

\(\displaystyle \lim_{x\rightarrow -\infty} 2x^2=2(-\infty)^2=2*\infty=\infty.\)

For infinity limits, we need only consider the leading term in both the numerator and the denominator.  Here, we have the case that the exponents are equal in the leading terms.  Therefore, the limit at infinity is simply the ratio of the coefficients of the leading terms. 

\(\displaystyle \lim_{x\rightarrow \infty}\frac{2x^2+3x}{10x^2+x}=\frac{2x^2}{10x^2}=\frac{2}{10}=\frac{1}{5}.\)

Infinity limits can be found by only considering the leading term in both the numerator and the denominator. In this problem, the numerator has a higher exponent than the denominator.  Therefore, it will keep increasing and increasing at a much faster rate.  These limits always tend to infinity.

\(\displaystyle \lim_{x\rightarrow \infty}\frac{3x^7+4x^4-3x^2}{5x^5+2x^4}=\frac{3x^7}{5x^5}=\frac{3}{5}x^2.\)

\(\displaystyle \lim_{x\rightarrow \infty}\frac{3}{5}x^2=\infty.\)

For infinity limits, we only consider the leading term in both the numerator and the denominator.  Then, we need to consider the exponents of the leading terms.  Here, the denominator has a higher degree than the numerator.  Therefore, we have a bottom heavy fraction.  Even though we are evaluating the limit at negative infinity, this will still tend to zero because the denominator is growing at a faster rate.  You can convince yourself of this by plugging in larger and larger negative values.  You will just get a longer and smaller decimal.

\(\displaystyle \lim_{x\rightarrow -\infty} \frac{2x^3-3x^2+4x}{4x^5-3x^3}=\frac{2x^3}{4x^5}=\frac{1}{2x^2}\)

\(\displaystyle \lim_{x\rightarrow -\infty}\frac{1}{2x^2}=0.\)

\(\displaystyle \begin{align*}&\text{Observation of the data points shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point centered approximately around:}\\&x=-1.8\\&\text{We find a zero denominator for the function: }\\&-\frac{(10x + 11)}{(11x + 20)}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Notice that as x increases, the points of data graphed appears}\\&\text{to level out, flattening towards a certain value. This value}\\&\text{is what is known as a horizontal asymptote. An asymptote is}\\&\text{a value that a function approaches, but never actually reaches.}\\&\text{Think of a horizontal asymptote as the limit of a function as}\\&\text{x approaches infinity. In such a case, as x approaches infinity,}\\&\text{any constants added or subtracted in the numerator and denominator}\\&\text{become irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=0.75\\&\text{This matches the ratio of coefficients for the function: }\\&\frac{(6x + 1)}{(8x - 17)}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Observation of the data points shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point centered approximately around:}\\&x=-1.1\\&\text{We find a zero denominator for the function: }\\&\frac{(8x - 2)}{(10x + 11)}\end{align*}\)