# Help me solve this rational problem?

## How do I solve this

Aug 9, 2018

See explanation.

#### Explanation:

Given: $\frac{{x}^{2} - 16}{{x}^{2} + x - 20}$

There are several things to note.

$\textcolor{b r o w n}{\text{Consider the numerator:}}$

${x}^{2} - 16 \to {a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Thus we have $\left(x - 4\right) \left(x + 4\right)$

From this it MAY be the case that the question designer intended one of them to cancel out in the denominator.

$\textcolor{b r o w n}{\text{Consider the denominator:}}$

As it is possible that there MAY be a cancelling out it is reasonable to explore $x = \pm 4$ as a factor

Note that $4 \times 5 = 20 \mathmr{and} 5 - 4 = 1$ so lets look at

${x}^{2} + x - 20 \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} \left(x + 5\right) \left(x - 4\right)$

color(white)("dddddddddddd")->color(white)("ddd") x^2+5x-4x-20 larr" Works"
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$\textcolor{b r o w n}{\text{Putting it all back together}}$

$\frac{{x}^{2} - 16}{{x}^{2} + x - 20} \to \frac{\cancel{\left(x - 4\right)} \left(x + 4\right)}{\left(x + 5\right) \cancel{\left(x - 4\right)}} = \frac{x + 4}{x + 5} \to f \left(x\right)$

As the equation is undefined when the denominator becomes 0 we have a vertical asymptote at $x = - 5$

${\lim}_{x \to + \infty} f \left(x\right) \to k = + 1$

${\lim}_{x \to - \infty} f \left(x\right) \to k = = + 1$

Thus there is a horizontal asymptote at $y = 1$
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$\textcolor{b r o w n}{\text{Any excluded value}}$

Although the $\left(x - 4\right)$ in the denominator cancels 'away' it still forms part of the original expression. Thus it has to be taken into account. This is done by setting $\left(x - 4\right) = 0$ and declaring $x = 4$ as the excluded value.
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$\textcolor{b r o w n}{\text{The general behaviour of the plot}}$

I will let you investigate this. You will need to shoe your logic in your solution.

We already know that there is a horizontal asymptote at $y = 1$

Test out what happens very close to the vertical asymptote. color(white)("d")

Aug 9, 2018

$\text{see explanation}$

#### Explanation:

$\text{let } f \left(x\right) = \frac{{x}^{2} - 16}{{x}^{2} + x - 20}$

$\text{the numerator is a "color(blue)"difference of squares}$

${x}^{2} - 16 = \left(x - 4\right) \left(x + 4\right)$

$\text{denominator}$

$\text{the factors of "-20" which sum to } + 1$
$\text{are "+5" and } - 4$

${x}^{2} + x - 20 = \left(x + 5\right) \left(x - 4\right)$

$f \left(x\right) = \frac{\cancel{\left(x - 4\right)} \left(x + 4\right)}{\left(x + 5\right) \cancel{\left(x - 4\right)}} = \frac{x + 4}{x + 5}$

$\text{the removal of the factor "(x-4)" from the numerator}$

$\text{ and denominator indicates a hole at}$

$x - 4 = 0 \Rightarrow x = 4 \to y = \frac{4 + 4}{4 + 5} = \frac{8}{9}$

$\text{hole at } \left(4 , \frac{8}{9}\right)$

$\text{the graph of "f(x)=(x+4)/(x+5)" is the same as}$

$\frac{{x}^{2} - 16}{{x}^{2} + x - 20} \text{ but without the hole}$

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.

$\text{solve "x+5=0rArrx=-5" is the asymptote}$

$\text{Horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by } x$

$f \left(x\right) = \frac{\frac{x}{x} + \frac{4}{x}}{\frac{x}{x} + \frac{5}{x}} = \frac{1 + \frac{4}{x}}{1 + \frac{5}{x}}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{1 + 0}$

$y = 1 \text{ is the asymptote}$

$\text{there is a discontinuity at } x = - 5$

$\text{domain is } x \in \left(- \infty , - 5\right) \cup \left(- 5 , \infty\right)$

$\text{there is a discontinuity at } y = 1$

$\text{range is } y \in \left(- \infty , 1\right) \cup \left(1 , \infty\right)$

$\text{For Intercepts}$

$x = 0 \Rightarrow y = \frac{4}{5} \leftarrow \textcolor{red}{\text{y-intercept}}$

$x + 4 = 0 \Rightarrow x = - 4 \leftarrow \textcolor{red}{\text{x-intercept}}$
graph{(x+4)/(x+5) [-10, 10, -5, 5]}