Question

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

Answer 1

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.

& \Rightarrow 0=\dfrac{8}{{{x}^{2}}-x-6} \\\ & \Rightarrow 0=8 \\\ \end{aligned}$$ We can see that the above step is not true and the x-intercept of the curve doesn’t lie. We know that at y-intercept, x = 0. $$\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 $$.