
How do you graph $y = \dfrac{{x + 2}}{{x + 3}}$ using asymptotes, intercepts, end behaviour?
Answer
559.8k+ views
Hint: In this question, we have a given equation and plotting the graphs of it with asymptotes will require finding its vertical and horizontal asymptotes. Also, to find intercepts, we will have to assume the values of one of the variables as 0, and find the other variable’s value. We will check the positive and the negative numbers in order to find the end behaviour.
Complete step-by-step answer:
We have an equation$y = \dfrac{{x + 2}}{{x + 3}}$, where we will first assume the value as $0$ for $x$,
$ \Rightarrow y = \dfrac{{0 + 2}}{{0 + 3}}$
Adding the numbers on the constant side, we get,
$ \Rightarrow y = \dfrac{2}{3}$
Therefore, $y$-intercept is $\dfrac{2}{3}$
Assuming$y = 0$,
$ \Rightarrow 0 = \dfrac{{x + 2}}{{x + 3}}$
Shifting the denominator to the left-hand side,
$ \Rightarrow (x + 3)0 = x + 2$
So,
$ \Rightarrow x = - 2$
Vertical asymptote at \[x = - 3\]
\[\begin{array}{*{20}{l}}
{y < 1\;when\;x > - 3} \\
{y > 1\;when\;x < - 3}
\end{array}\]
Explanation:
Asymptote:
$\dfrac{n}{0}$ = undefined
If \[x + 3 = 0\], \[y\] is undefined.
This means that it is not on the graph, and so is shown as the asymptote.
When \[x + 3 = 0,\;x = 0 - 3\]
So,
\[ \Rightarrow x - 3\]
Intercepts:
\[y:\]
The \[y\]-intercept is when \[x = 0\]
\[ \Rightarrow y = x + 2x + 3\]
So now,
$ \Rightarrow y = 23$
$y$-intercept is \[23\]
$x$:
The $x$-intercept is when \[y = 0\]
\[ \Rightarrow x + 2x + 3 = 0\]
The numerator has to be $0$, since $\dfrac{0}{n}$=$0$
This means that \[x + 2 = 0\]
When \[x + 2 = 0,\;x = 0 - 2\]
$x$-intercept is $ - 2$
End behaviour:
\[y < 1\]
\[x + 2\;\] is always less than \[x + 3\]. with both positive and negative numbers.
The fraction \[\dfrac{{x + 2}}{{x + 3}}\] cannot be simplified \[to\;\dfrac{{ \geqslant 1}}{1}\] if $x > - 3$
However, it is vice versa for negative numbers, since smaller negative numbers have a higher absolute value (distance from 0),
This means that \[x + 2x + 3\;\] cannot be simplified \[to\;\dfrac{{ \geqslant 1}}{1}\] if $x > - 3$
Note:
In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tend to infinity. In projective geometry and related contexts, an asymptote of a curve is a line which is tangent to the curve at a point at infinity. In analytic geometry, using the common convention that the horizontal axis represents a variable x and vertical axis represents a variable y, a y-intercept or vertical intercept is a point where the graph of a function or relation intersects the y-axis of the coordinate system. The end behaviour of a function f describes the behaviour of the graph of the function at the “ends” of the x-axis.
Complete step-by-step answer:
We have an equation$y = \dfrac{{x + 2}}{{x + 3}}$, where we will first assume the value as $0$ for $x$,
$ \Rightarrow y = \dfrac{{0 + 2}}{{0 + 3}}$
Adding the numbers on the constant side, we get,
$ \Rightarrow y = \dfrac{2}{3}$
Therefore, $y$-intercept is $\dfrac{2}{3}$
Assuming$y = 0$,
$ \Rightarrow 0 = \dfrac{{x + 2}}{{x + 3}}$
Shifting the denominator to the left-hand side,
$ \Rightarrow (x + 3)0 = x + 2$
So,
$ \Rightarrow x = - 2$
Vertical asymptote at \[x = - 3\]
\[\begin{array}{*{20}{l}}
{y < 1\;when\;x > - 3} \\
{y > 1\;when\;x < - 3}
\end{array}\]
Explanation:
Asymptote:
$\dfrac{n}{0}$ = undefined
If \[x + 3 = 0\], \[y\] is undefined.
This means that it is not on the graph, and so is shown as the asymptote.
When \[x + 3 = 0,\;x = 0 - 3\]
So,
\[ \Rightarrow x - 3\]
Intercepts:
\[y:\]
The \[y\]-intercept is when \[x = 0\]
\[ \Rightarrow y = x + 2x + 3\]
So now,
$ \Rightarrow y = 23$
$y$-intercept is \[23\]
$x$:
The $x$-intercept is when \[y = 0\]
\[ \Rightarrow x + 2x + 3 = 0\]
The numerator has to be $0$, since $\dfrac{0}{n}$=$0$
This means that \[x + 2 = 0\]
When \[x + 2 = 0,\;x = 0 - 2\]
$x$-intercept is $ - 2$
End behaviour:
\[y < 1\]
\[x + 2\;\] is always less than \[x + 3\]. with both positive and negative numbers.
The fraction \[\dfrac{{x + 2}}{{x + 3}}\] cannot be simplified \[to\;\dfrac{{ \geqslant 1}}{1}\] if $x > - 3$
However, it is vice versa for negative numbers, since smaller negative numbers have a higher absolute value (distance from 0),
This means that \[x + 2x + 3\;\] cannot be simplified \[to\;\dfrac{{ \geqslant 1}}{1}\] if $x > - 3$
Note:
In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tend to infinity. In projective geometry and related contexts, an asymptote of a curve is a line which is tangent to the curve at a point at infinity. In analytic geometry, using the common convention that the horizontal axis represents a variable x and vertical axis represents a variable y, a y-intercept or vertical intercept is a point where the graph of a function or relation intersects the y-axis of the coordinate system. The end behaviour of a function f describes the behaviour of the graph of the function at the “ends” of the x-axis.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

