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How do you find the domain, points of discontinuity and x and y intercepts of the rational function \[y = \dfrac{{12 - 6x}}{{{x^2} - 8x + 12}}\]?

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Last updated date: 13th Jun 2024
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Answer
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Hint: First factorize the numerator and denominator. Put the denominator to zero to find the point of discontinuity and also remove those points from the domain. For x intercept, put $y = 0$ and for y intercept, put $x = 0$ in the function.

Complete step by step answer:
According to the question, we have been given a rational function and we have to determine its domain, points of discontinuity and its x and y intercepts.
The function is:
\[ \Rightarrow y = \dfrac{{12 - 6x}}{{{x^2} - 8x + 12}}\]
We can simplify the function by taking the common factor out in the numerator and factoring the quadratic expression in the denominator. So if we denote the denominator by $D$ and factorize it subsequently, we have:
$
   \Rightarrow D = {x^2} - 8x + 12 \\
   \Rightarrow D = {x^2} - 6x - 2x + 12 \\
   \Rightarrow D = x\left( {x - 6} \right) - 2\left( {x - 6} \right) \\
   \Rightarrow D = \left( {x - 2} \right)\left( {x - 6} \right) \\
 $
Putting this in the function and taking the common factor out in numerator, we’ll get:
\[ \Rightarrow y = \dfrac{{ - 6\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 6} \right)}}{\text{ }}.....{\text{(1)}}\]
Now we can easily determine the things asked in the question.
Domain:
We know that for a rational function to be defined, its denominator can’t be zero. So the domain of this function will be all real values except the points where its denominator is zero. The points where the denominator is zero is calculated below:
$
   \Rightarrow \left( {x - 2} \right)\left( {x - 6} \right) = 0 \\
   \Rightarrow \left( {x - 2} \right) = 0{\text{ and }}\left( {x - 6} \right) = 0 \\
   \Rightarrow x = 2{\text{ and }}x = 6 \\
 $
Therefore the domain of the function will not include $x = 2$ and $x = 6$.
$ \Rightarrow {\text{Domain}} \equiv R - \{ 3,6\} $
Points of discontinuity:
The function will also be discontinuous at the points where it is not defined. At all other points except those, the function will be continuous and will have a finite value. Thus the points of discontinuity of the are $x = 2$ and $x = 6$.
Intercepts:
We will find the x intercept by putting $y = 0$ in the function i.e. equation (1). Doing so, we’ll get:
\[
   \Rightarrow 0 = \dfrac{{ - 6\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 6} \right)}} \\
   \Rightarrow - 6\left( {x - 2} \right) = 0 \\
   \Rightarrow x = 2 \\
 \]
Similarly, y intercept can be obtained by putting $x = 0$ in the function. We’ll get:
\[
   \Rightarrow y = \dfrac{{ - 6\left( {0 - 2} \right)}}{{\left( {0 - 2} \right)\left( {0 - 6} \right)}} \\
   \Rightarrow y = \dfrac{{ - 6\left( { - 2} \right)}}{{\left( { - 2} \right)\left( { - 6} \right)}} \\
   \Rightarrow y = 1 \\
 \]

Therefore x intercept is $x = 2$ and y intercept is $y = 1$

Note: If we observe the function in equation (1) again, we have:
\[ \Rightarrow y = \dfrac{{ - 6\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 6} \right)}}\]
One important point to keep in mind here is that we can’t cancel out the common factors from the numerator and denominator before finding the domain and point of discontinuity else this will lead to incorrect results.
Once we have determined the domain and point of discontinuity then we can cancel the common factor to make it simpler.
\[ \Rightarrow y = \dfrac{{ - 6}}{{\left( {x - 6} \right)}}\]