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Question: How do you graph \[y=\dfrac{4}{3x-6}+5\] using asymptotes, intercepts, end behaviour?...

How do you graph y=43x6+5y=\dfrac{4}{3x-6}+5 using asymptotes, intercepts, end behaviour?

Explanation

Solution

These types of problems are somehow easy to solve once we have figured out the underlying concept behind the sum. This requires a strong knowledge of topics like functions, equations, continuity and graph theory. When we see these types of problems, the first thing that we need to do is to find out all the asymptotic points or lines. These are lines such that the graph intersects these lines at infinity. After that we need to find out the critical points of the graph, which can be done by differentiating the function once and then equating it to zero. We then need to find the nature of the graph, which is done by differentiating the given function two times and then equating it to zero.

Complete step by step solution:
Now we start off with the solution to the above given problem by finding out if there exists any asymptotes or not. We can clearly see that if x=2x=2 is put in the given equation, then the value of ‘y’ tends to infinity. So we can say x=2x=2 is an asymptote and the graph meets this line at infinity.
We can put x=0x=0 to find out the y-intercept, which comes out to be y=46+5y=-\dfrac{4}{6}+5
y=133\Rightarrow y=\dfrac{13}{3}
We can put y=0y=0 to find out the x-intercept, which comes out to be 43x6+5=0\dfrac{4}{3x-6}+5=0

& \Rightarrow \dfrac{4}{3x-6}+5=0 \\\ & \Rightarrow \dfrac{4}{3x-6}=-5 \\\ & \Rightarrow \dfrac{4}{5}=-\left( 3x-6 \right) \\\ & \Rightarrow \dfrac{4}{5}=-3x+6 \\\ & \Rightarrow 3x=6-\dfrac{4}{5} \\\ & \therefore x=\dfrac{26}{15} \\\ \end{aligned}$$ Now we differentiate the equation two times to get, $$\begin{aligned} & \dfrac{dy}{dx}=-\dfrac{12}{{{\left( 3x-6 \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{72}{{{\left( 3x-6 \right)}^{3}}} \\\ \end{aligned}$$ Now from this we can say that if $$x>2$$ then the second derivative of the function will be positive and the graph of the function will be concave up. If $$x<2$$ then the second derivative of the function will be negative and the graph of the function will be concave down. Thus keeping in mind of all these concepts, we plot the graph of the equation as, ![](https://www.vedantu.com/question-sets/271034e1-907e-4122-9d27-b107b5a6865b7658093490934575739.png) **Note:** For problems like these the most important thing is to have a fair idea of graph theory and its related terms. For graph theory, we need to have an idea of functions, continuity and differentiation. We need to find all the prerequisites that we need to have before we can plot bare hand. Plotting graphs are very helpful in solving various complex problems which become easy on doing so. The nature of the graph can easily be found out using the second differentiation of the given function.