Question

How do you graph \[y=\dfrac{8}{{{x}^{2}}-x-6}\] using asymptotes, intercepts, end behaviour?

Answer 1

Hint: In this problem, we are given an equation of a curve for which we have to draw a graph using asymptotes, intercepts and end behaviour. We can first write the given equation in the form of \[\dfrac{n\left( x \right)}{d\left( x \right)}\]. We can then find the asymptotes as the vertical asymptotes are obtained by \[y\to \infty \] and horizontal asymptotes are obtained by \[x\to \infty \]. We can then find the x-intercept where the value of y is zero and the y-intercept where the value of x is zero. To find the behaviour of the curve, we can substitute \[x=\infty ,y=\infty \].

Complete step by step solution:
We know that the given equation is,
\[y=\dfrac{8}{{{x}^{2}}-x-6}\]
We can write the above equation in the form of \[\dfrac{n\left( x \right)}{d\left( x \right)}\], we get
 \[\dfrac{n\left( x \right)}{d\left( x \right)}=\dfrac{8}{{{x}^{2}}-x-6}\]
Now we can find the vertical and the horizontal asymptote.
We know that vertical asymptotes are obtained by \[y\to \infty \].
We can see that when \[y\to \infty \], \[d\left( x \right)=0\], we can now substitute, we get
\[\Rightarrow {{x}^{2}}-x-6=0\]
We can now factorize the above step, we get
\[\Rightarrow \left( x-3 \right)\left( x+2 \right)=0\]
We can now equate the above step, we get
\[\Rightarrow x=3,-2\]
The vertical asymptote of the curve is at x = 3 and x = -2.
Now we can determine the horizontal asymptote as \[x\to \infty \], we get
\[\Rightarrow y=\dfrac{8}{{{\infty }^{2}}-\infty -6}=\dfrac{8}{\infty }=0\]
So, the horizontal asymptote of the curve is at y = 0.
We can now find the x-intercept and the y-intercept.
We know that at x-intercept, y = 0.
\[\begin{align}
  & \Rightarrow 0=\dfrac{8}{{{x}^{2}}-x-6} \\
 & \Rightarrow 0=8 \\
\end{align}\]
We can see that the above step is not true and the x-intercept of the curve doesn’t lie.
We know that at y-intercept, x = 0.
\[\Rightarrow y=\dfrac{8}{{{0}^{2}}-0-6}=-\dfrac{4}{3}\]
Therefore, the y-intercept of the curve is at \[\left( 0,-\dfrac{4}{3} \right)\].
We can now find the behaviour of the curve.
We can now take \[x\to \infty \], then
\[\Rightarrow y=\dfrac{8}{{{\infty }^{2}}-\infty -6}=0\]
We can now take \[x\to -\infty \], the
\[\Rightarrow y=\dfrac{8}{{{\left( -\infty \right)}^{2}}-\left( -\infty \right)-6}=0\]
This means as x approaches towards infinity, the curve touches the x axis.
So, the end behaviour is,
As \[x\to \infty \],\[y\to \infty \]
As \[x\to -\infty \],\[y\to \infty \]
Therefore, the vertical asymptote of the curve is at x = 3 and x = -2, the horizontal asymptote of the curve is at y = 0. The y-intercept of the curve is at \[\left( 0,-\dfrac{4}{3} \right)\], the x-intercept of the curve doesn’t lie.
The end behaviour is,
As \[x\to \infty \],\[y\to \infty \]
As \[x\to -\infty \],\[y\to \infty \]
We can now draw the graph for the above values.

Note: We should remember that asymptotes of the curve is a line such that the distance between the line and the curve approaches towards zero as one or both of the x or y coordinates tends to infinity. We should also remember that as the vertical asymptotes are obtained by \[y\to \infty \] and horizontal asymptotes are obtained by \[x\to \infty \].