How do you write an equation with a vertical asymptote of \[3\], slant asymptote of \[y=x+1\], and \[x\] intercept at \[2\]?
Answer
605.4k+ views
Hint: From the question we have been asked to write the equation with the given vertical and slant asymptote along with \[x\] intercept. For solving this question we will assume that the required equation be \[f(x)\] as a fractional function as we are given the vertical asymptote. Then we will use the slant asymptote given and simplify our \[f(x)\] equation and extract our required solution.
Complete step-by-step solution:
Firstly, let us assume that \[f(x)=\dfrac{n(x)}{d(x)}\] be a fractional function as said before.
We are given that it has a vertical asymptote a \[3\].so, we get that,
\[\Rightarrow d(x)=(x-3)\]
We are also given that slant asymptote of \[y=x+1\], so we got to know that the degree numerator function that is \[n(x)\] must be two.
Now, we use the general equation of a polynomial. We know that the general polynomial of degree two is as follows.
\[\Rightarrow n(x)=a{{x}^{2}}+bx+c\]
So, we now substitute these two in our function \[f(x)=\dfrac{n(x)}{d(x)}\]. So, we get the equation reduced as follows.
\[\Rightarrow f(x)=\dfrac{a{{x}^{2}}+bx+c}{x-3}\]
We also know that
\[\Rightarrow f(2)=0\] and \[a{{x}^{2}}+bx+c=\left( x-3 \right)\left( x+1 \right)+d\] because,
\[\Rightarrow f(x)=\dfrac{a{{x}^{2}}+bx+c}{x-3}=(x+1)+\dfrac{d}{x-3}\]
So, for big values of \[\left| x \right|\]
\[\Rightarrow f(x)\approx (x+1)\]
Then we get \[a{{x}^{2}}+bx+c={{x}^{2}}-2x-3+d\]
So, by comparing we got the values as follows.
\[\Rightarrow a=1,b=-2,c=d-3\]
Now, we will substitute the \[\Rightarrow f(2)=0\]. So, from this we extract the following.
\[\Rightarrow 4a+2b+c=0\]
Now, we substitute the values of a,b,c we got in the above equation.
\[\Rightarrow 4-4+d-3=0\]
\[\Rightarrow d=3\]
Therefore, the equation will be \[\Rightarrow f(x)=(x+1)+\dfrac{-3}{x-3}=\dfrac{{{x}^{2}}-2x}{x-3}\].
The graph of the function will be as follows.
Note: Students must be able to do the calculations accurately. Students should have good concepts in functions and they must know the definitions and applications of the asymptotes, intercepts in the concept of functions. We must use the standard form of second degree equation which is \[\Rightarrow n(x)=a{{x}^{2}}+bx+c\] to solve the question.
Complete step-by-step solution:
Firstly, let us assume that \[f(x)=\dfrac{n(x)}{d(x)}\] be a fractional function as said before.
We are given that it has a vertical asymptote a \[3\].so, we get that,
\[\Rightarrow d(x)=(x-3)\]
We are also given that slant asymptote of \[y=x+1\], so we got to know that the degree numerator function that is \[n(x)\] must be two.
Now, we use the general equation of a polynomial. We know that the general polynomial of degree two is as follows.
\[\Rightarrow n(x)=a{{x}^{2}}+bx+c\]
So, we now substitute these two in our function \[f(x)=\dfrac{n(x)}{d(x)}\]. So, we get the equation reduced as follows.
\[\Rightarrow f(x)=\dfrac{a{{x}^{2}}+bx+c}{x-3}\]
We also know that
\[\Rightarrow f(2)=0\] and \[a{{x}^{2}}+bx+c=\left( x-3 \right)\left( x+1 \right)+d\] because,
\[\Rightarrow f(x)=\dfrac{a{{x}^{2}}+bx+c}{x-3}=(x+1)+\dfrac{d}{x-3}\]
So, for big values of \[\left| x \right|\]
\[\Rightarrow f(x)\approx (x+1)\]
Then we get \[a{{x}^{2}}+bx+c={{x}^{2}}-2x-3+d\]
So, by comparing we got the values as follows.
\[\Rightarrow a=1,b=-2,c=d-3\]
Now, we will substitute the \[\Rightarrow f(2)=0\]. So, from this we extract the following.
\[\Rightarrow 4a+2b+c=0\]
Now, we substitute the values of a,b,c we got in the above equation.
\[\Rightarrow 4-4+d-3=0\]
\[\Rightarrow d=3\]
Therefore, the equation will be \[\Rightarrow f(x)=(x+1)+\dfrac{-3}{x-3}=\dfrac{{{x}^{2}}-2x}{x-3}\].
The graph of the function will be as follows.
Note: Students must be able to do the calculations accurately. Students should have good concepts in functions and they must know the definitions and applications of the asymptotes, intercepts in the concept of functions. We must use the standard form of second degree equation which is \[\Rightarrow n(x)=a{{x}^{2}}+bx+c\] to solve the question.
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