Question

How do you graph $f\left( x \right)=\dfrac{{{x}^{2}}+3x}{-4x+4}$ using holes, vertical and horizontal asymptotes, $x$ and $y$ intercepts ?

Answer 1

To find the holes , we check at which values of $x$ , the given function is undefined. An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both the coordinates tends to infinity. To find the vertical asymptotes, we equate the denominator to $0$. To find the horizontal asymptotes, we compare the degree of the numerator and the degree of the denominator. To find the $x$ and $y$ intercepts, substitute $f\left( x \right)=0$ and $x=0$ respectively.

Complete step by step solution:
Holes in the graph of a rational function are the points on the curve of the function where the value of the function is not defined.
In a rational function, the value of the function $f\left( x \right)$ is undefined when either both the numerator and the denominator are zero or both of them tend to infinity.
The given function is $f\left( x \right)=\dfrac{{{x}^{2}}+3x}{-4x+4}$.
Let us equate the numerator and the denominator to $0$ to see if there are being equal to $0$ for only one value of $x$.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow {{x}^{2}}+3x=0 \\\ & \Rightarrow x\left( x+3 \right)=0 \\\ & \Rightarrow x=0,x=-3 \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow -4x+4=0 \\\ & \Rightarrow -4\left( x-1 \right)=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}$
We can clearly see that for only value of $x$ the numerator and the denominator are not equal to $0$. We can conclude that there are no holes for this function.
Vertical asymptote is the line that becomes a vertical tangent to the curve at infinite value of $f\left( x \right)$ . We can find the equation of the vertical asymptotes by equating the denominator to zero.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow -4x+4=0 \\\ & \Rightarrow -4\left( x-1 \right)=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}$
This means that the vertical asymptote of the graph is $x=1$.
Horizontal asymptote is the tangent to the curve at infinite value of $x$. When the degree of the denominator is greater or equal to the degree of the numerator, then the horizontal asymptote is equal to the ratio of the coefficients of the terms with the highest power in the denominator and the numerator.
But if the degree of the numerator is greater than the degree of the denominator , then that function has no horizontal asymptote.
This function has no horizontal asymptote.
$x$ intercepts are the points where the curve cuts or touches $x$-axis. We can find thee points by equating the function to zero.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow \dfrac{{{x}^{2}}+3x}{-4x+4}=0 \\\ & \Rightarrow {{x}^{2}}+3x=0 \\\ & \Rightarrow x\left( x+3 \right)=0 \\\ & \Rightarrow x=0,x=-3 \\\ \end{aligned}$
$x$ intercepts is $x=0,x=-3$.
$y$ intercepts are the points where the curve cuts or touches $y$-axis. We can find thee points by substituting $x=0$.
$\begin{aligned} & \Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}+3x}{-4x+4} \\\ & \Rightarrow f\left( 0 \right)=\dfrac{{{\left( 0 \right)}^{2}}+3\left( 0 \right)}{-4\left( 0 \right)+4}=0 \\\ \end{aligned}$
Now let us use all this information and graph it.
Graph :

Note: We should find out vertical asymptotes, horizontal asymptotes, holes , intercepts. We should know their definitions. We should have a lot of practice in graphing difficult functions so as to be able to graph a given tough function in the exam. Sometimes, we are also asked to trace the graph. We trace a function by using the first and second rule of differentiation. We also check whether the graph is identical with respect to $x/y$ axis or both the axes.