Mixing
Understanding graph of this rational function
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have this equation: $(4 - 3x - x^2)/(x - 3)$ and I am learning Pre-Calculus from coolmath and using desmos to graph.
I can easily calculate:
Y-intercept: $f(0) = (0,-4/3)$
X-intercepts: $4 - 3x - x^2 = 0$ ==> $(1,0)$ and $(-4,0)$.
Vertical Asymptote: $x - 3 = 0$ => $(3,0)$
No Horizontal Asymptote
Oblique Asymptote: $(4 - 3x - x^2)/(x - 3)$ ==> $-x - 6$
Graph came out to be like this:
Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)
Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.
graphing-functions rational-functions
edited Jul 21 at 15:37
mvw
30.4k22250
asked Jul 21 at 14:59
Arnuld
4510
add a comment |Â
up vote
0
down vote
favorite
I have this equation: $(4 - 3x - x^2)/(x - 3)$ and I am learning Pre-Calculus from coolmath and using desmos to graph.
I can easily calculate:
Y-intercept: $f(0) = (0,-4/3)$
X-intercepts: $4 - 3x - x^2 = 0$ ==> $(1,0)$ and $(-4,0)$.
Vertical Asymptote: $x - 3 = 0$ => $(3,0)$
No Horizontal Asymptote
Oblique Asymptote: $(4 - 3x - x^2)/(x - 3)$ ==> $-x - 6$
Graph came out to be like this:
Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)
Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.
graphing-functions rational-functions
edited Jul 21 at 15:37
mvw
30.4k22250
asked Jul 21 at 14:59
Arnuld
4510
1
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have this equation: $(4 - 3x - x^2)/(x - 3)$ and I am learning Pre-Calculus from coolmath and using desmos to graph.
I can easily calculate:
Y-intercept: $f(0) = (0,-4/3)$
X-intercepts: $4 - 3x - x^2 = 0$ ==> $(1,0)$ and $(-4,0)$.
Vertical Asymptote: $x - 3 = 0$ => $(3,0)$
No Horizontal Asymptote
Oblique Asymptote: $(4 - 3x - x^2)/(x - 3)$ ==> $-x - 6$
Graph came out to be like this:
Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)
Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.
graphing-functions rational-functions
edited Jul 21 at 15:37
mvw
30.4k22250
asked Jul 21 at 14:59
Arnuld
4510
I have this equation: $(4 - 3x - x^2)/(x - 3)$ and I am learning Pre-Calculus from coolmath and using desmos to graph.
I can easily calculate:
Y-intercept: $f(0) = (0,-4/3)$
X-intercepts: $4 - 3x - x^2 = 0$ ==> $(1,0)$ and $(-4,0)$.
Vertical Asymptote: $x - 3 = 0$ => $(3,0)$
No Horizontal Asymptote
Oblique Asymptote: $(4 - 3x - x^2)/(x - 3)$ ==> $-x - 6$
Graph came out to be like this:
Problem #1: I thought graph has 3 parts but it got 2 (there is nothing in the middle)
Problem #2: How to deduce the shape of the graph from all these calculations. I mean the the turns and curves and directions without plotting the points. With points anyone can plot it but without plotting, it requires some Mathematical-Thinking and that is the thing I want to develop.
graphing-functions rational-functions
edited Jul 21 at 15:37
mvw
30.4k22250
asked Jul 21 at 14:59
Arnuld
4510
edited Jul 21 at 15:37
mvw
30.4k22250
edited Jul 21 at 15:37
mvw
30.4k22250
edited Jul 21 at 15:37
mvw
30.4k22250
30.4k22250
asked Jul 21 at 14:59
Arnuld
4510
asked Jul 21 at 14:59
Arnuld
4510
asked Jul 21 at 14:59
Arnuld
4510
4510
1
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27
add a comment |Â
1
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27
1
1
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Regarding your first problem:
You can guess the function has "2 parts" because it is continuous everywhere except $x=3$. so you know the line $x=3$ will divide the 2 parts where the function is continuous.
Regarding your second problem:
When graphing the function, look at:
- points of discontinuity
- roots of the functions (where the function equals 0)
- minimum and maximum
- notice the numerator is of degree 2 and the denominator is of degree 1, so the function acts like a linear function as x goes further from the origin
answered Jul 21 at 15:22
GuySa
402313
add a comment |Â
up vote
0
down vote
Your function
$$
y = (4−3x−x^2)/(x−3)
$$
Can be rewritten as
beginalign
0
&= (x-3) y + x^2 + 3x - 4 \
&= xy - 3y + x^2 + 3x - 4 \
&= x^2 + xy + 3x - 3y - 4 \
endalign
which is an instance of a conic section
$$
A x^2 + B xy + Cy^2 + D x + E y + F = 0
$$
This is an ancient study subject. There is much known how the coefficients will select what solution curves.
In your case it is a hyperbola.
answered Jul 21 at 15:47
mvw
30.4k22250
add a comment |Â
up vote
0
down vote
What you were doing there is some kind of curve sketiching but without the powerful tools of calculus itself. But the things you denoted there are quite useful aswell.
First of all you know something about the asymptotic behavior of the function in two ways. So you can just say that the function will go to infinity while the x-value is decreasing and it will go to negative infinity while the x-value is increasing. These you can just derivate from the oblique asymptote you figured out. So now you know how the function will develop with the x argument.
As a next step concentrate on the oblique asymptote. You know the y-value will reach infinity in the near of this asymptote. When you put this together with the roots of the polynomial you can go further with the form of the graph.
Both roots are smaller than the x-value of the oblique asymptote. Also the function goes for infinity for big negative x-values. So the graph will fall while you going to the first root but right after this it have to rise against since the graph has to reach the next root AND has to go for infinity at the oblique asymptote.
On the other hand the right side of the graph never hits the x-axis hence it goes from negative infinity up while the x-value is decreasing. So if the function would now go for infinity again there would be another interception with the x-axis. So it goes to negative infinity.
To put it all together: From the asymptotes you know how the function behaves at critical points such as infinity, negative infinity and at poles. Now you just focus on the values of the roots and compare them with the poles. From there on you can sketch the graph in a way of thinking how it develops at several x-values.
answered Jul 21 at 15:30
mrtaurho
700219
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Regarding your first problem:
You can guess the function has "2 parts" because it is continuous everywhere except $x=3$. so you know the line $x=3$ will divide the 2 parts where the function is continuous.
Regarding your second problem:
When graphing the function, look at:
- points of discontinuity
- roots of the functions (where the function equals 0)
- minimum and maximum
- notice the numerator is of degree 2 and the denominator is of degree 1, so the function acts like a linear function as x goes further from the origin
answered Jul 21 at 15:22
GuySa
402313
add a comment |Â
up vote
1
down vote
Regarding your first problem:
You can guess the function has "2 parts" because it is continuous everywhere except $x=3$. so you know the line $x=3$ will divide the 2 parts where the function is continuous.
Regarding your second problem:
When graphing the function, look at:
- points of discontinuity
- roots of the functions (where the function equals 0)
- minimum and maximum
- notice the numerator is of degree 2 and the denominator is of degree 1, so the function acts like a linear function as x goes further from the origin
answered Jul 21 at 15:22
GuySa
402313
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Regarding your first problem:
You can guess the function has "2 parts" because it is continuous everywhere except $x=3$. so you know the line $x=3$ will divide the 2 parts where the function is continuous.
Regarding your second problem:
When graphing the function, look at:
- points of discontinuity
- roots of the functions (where the function equals 0)
- minimum and maximum
- notice the numerator is of degree 2 and the denominator is of degree 1, so the function acts like a linear function as x goes further from the origin
answered Jul 21 at 15:22
GuySa
402313
Regarding your first problem:
You can guess the function has "2 parts" because it is continuous everywhere except $x=3$. so you know the line $x=3$ will divide the 2 parts where the function is continuous.
Regarding your second problem:
When graphing the function, look at:
- points of discontinuity
- roots of the functions (where the function equals 0)
- minimum and maximum
- notice the numerator is of degree 2 and the denominator is of degree 1, so the function acts like a linear function as x goes further from the origin
answered Jul 21 at 15:22
GuySa
402313
edited Jul 21 at 15:28
answered Jul 21 at 15:22
GuySa
402313
answered Jul 21 at 15:22
GuySa
402313
answered Jul 21 at 15:22
GuySa
402313
402313
add a comment |Â
add a comment |Â
up vote
0
down vote
Your function
$$
y = (4−3x−x^2)/(x−3)
$$
Can be rewritten as
beginalign
0
&= (x-3) y + x^2 + 3x - 4 \
&= xy - 3y + x^2 + 3x - 4 \
&= x^2 + xy + 3x - 3y - 4 \
endalign
which is an instance of a conic section
$$
A x^2 + B xy + Cy^2 + D x + E y + F = 0
$$
This is an ancient study subject. There is much known how the coefficients will select what solution curves.
In your case it is a hyperbola.
answered Jul 21 at 15:47
mvw
30.4k22250
add a comment |Â
up vote
0
down vote
Your function
$$
y = (4−3x−x^2)/(x−3)
$$
Can be rewritten as
beginalign
0
&= (x-3) y + x^2 + 3x - 4 \
&= xy - 3y + x^2 + 3x - 4 \
&= x^2 + xy + 3x - 3y - 4 \
endalign
which is an instance of a conic section
$$
A x^2 + B xy + Cy^2 + D x + E y + F = 0
$$
This is an ancient study subject. There is much known how the coefficients will select what solution curves.
In your case it is a hyperbola.
answered Jul 21 at 15:47
mvw
30.4k22250
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your function
$$
y = (4−3x−x^2)/(x−3)
$$
Can be rewritten as
beginalign
0
&= (x-3) y + x^2 + 3x - 4 \
&= xy - 3y + x^2 + 3x - 4 \
&= x^2 + xy + 3x - 3y - 4 \
endalign
which is an instance of a conic section
$$
A x^2 + B xy + Cy^2 + D x + E y + F = 0
$$
This is an ancient study subject. There is much known how the coefficients will select what solution curves.
In your case it is a hyperbola.
answered Jul 21 at 15:47
mvw
30.4k22250
Your function
$$
y = (4−3x−x^2)/(x−3)
$$
Can be rewritten as
beginalign
0
&= (x-3) y + x^2 + 3x - 4 \
&= xy - 3y + x^2 + 3x - 4 \
&= x^2 + xy + 3x - 3y - 4 \
endalign
which is an instance of a conic section
$$
A x^2 + B xy + Cy^2 + D x + E y + F = 0
$$
This is an ancient study subject. There is much known how the coefficients will select what solution curves.
In your case it is a hyperbola.
answered Jul 21 at 15:47
mvw
30.4k22250
answered Jul 21 at 15:47
mvw
30.4k22250
answered Jul 21 at 15:47
mvw
30.4k22250
answered Jul 21 at 15:47
mvw
30.4k22250
30.4k22250
add a comment |Â
add a comment |Â
up vote
0
down vote
What you were doing there is some kind of curve sketiching but without the powerful tools of calculus itself. But the things you denoted there are quite useful aswell.
First of all you know something about the asymptotic behavior of the function in two ways. So you can just say that the function will go to infinity while the x-value is decreasing and it will go to negative infinity while the x-value is increasing. These you can just derivate from the oblique asymptote you figured out. So now you know how the function will develop with the x argument.
As a next step concentrate on the oblique asymptote. You know the y-value will reach infinity in the near of this asymptote. When you put this together with the roots of the polynomial you can go further with the form of the graph.
Both roots are smaller than the x-value of the oblique asymptote. Also the function goes for infinity for big negative x-values. So the graph will fall while you going to the first root but right after this it have to rise against since the graph has to reach the next root AND has to go for infinity at the oblique asymptote.
On the other hand the right side of the graph never hits the x-axis hence it goes from negative infinity up while the x-value is decreasing. So if the function would now go for infinity again there would be another interception with the x-axis. So it goes to negative infinity.
To put it all together: From the asymptotes you know how the function behaves at critical points such as infinity, negative infinity and at poles. Now you just focus on the values of the roots and compare them with the poles. From there on you can sketch the graph in a way of thinking how it develops at several x-values.
answered Jul 21 at 15:30
mrtaurho
700219
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
add a comment |Â
up vote
0
down vote
What you were doing there is some kind of curve sketiching but without the powerful tools of calculus itself. But the things you denoted there are quite useful aswell.
First of all you know something about the asymptotic behavior of the function in two ways. So you can just say that the function will go to infinity while the x-value is decreasing and it will go to negative infinity while the x-value is increasing. These you can just derivate from the oblique asymptote you figured out. So now you know how the function will develop with the x argument.
As a next step concentrate on the oblique asymptote. You know the y-value will reach infinity in the near of this asymptote. When you put this together with the roots of the polynomial you can go further with the form of the graph.
Both roots are smaller than the x-value of the oblique asymptote. Also the function goes for infinity for big negative x-values. So the graph will fall while you going to the first root but right after this it have to rise against since the graph has to reach the next root AND has to go for infinity at the oblique asymptote.
On the other hand the right side of the graph never hits the x-axis hence it goes from negative infinity up while the x-value is decreasing. So if the function would now go for infinity again there would be another interception with the x-axis. So it goes to negative infinity.
To put it all together: From the asymptotes you know how the function behaves at critical points such as infinity, negative infinity and at poles. Now you just focus on the values of the roots and compare them with the poles. From there on you can sketch the graph in a way of thinking how it develops at several x-values.
answered Jul 21 at 15:30
mrtaurho
700219
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What you were doing there is some kind of curve sketiching but without the powerful tools of calculus itself. But the things you denoted there are quite useful aswell.
First of all you know something about the asymptotic behavior of the function in two ways. So you can just say that the function will go to infinity while the x-value is decreasing and it will go to negative infinity while the x-value is increasing. These you can just derivate from the oblique asymptote you figured out. So now you know how the function will develop with the x argument.
As a next step concentrate on the oblique asymptote. You know the y-value will reach infinity in the near of this asymptote. When you put this together with the roots of the polynomial you can go further with the form of the graph.
Both roots are smaller than the x-value of the oblique asymptote. Also the function goes for infinity for big negative x-values. So the graph will fall while you going to the first root but right after this it have to rise against since the graph has to reach the next root AND has to go for infinity at the oblique asymptote.
On the other hand the right side of the graph never hits the x-axis hence it goes from negative infinity up while the x-value is decreasing. So if the function would now go for infinity again there would be another interception with the x-axis. So it goes to negative infinity.
To put it all together: From the asymptotes you know how the function behaves at critical points such as infinity, negative infinity and at poles. Now you just focus on the values of the roots and compare them with the poles. From there on you can sketch the graph in a way of thinking how it develops at several x-values.
answered Jul 21 at 15:30
mrtaurho
700219
What you were doing there is some kind of curve sketiching but without the powerful tools of calculus itself. But the things you denoted there are quite useful aswell.
First of all you know something about the asymptotic behavior of the function in two ways. So you can just say that the function will go to infinity while the x-value is decreasing and it will go to negative infinity while the x-value is increasing. These you can just derivate from the oblique asymptote you figured out. So now you know how the function will develop with the x argument.
As a next step concentrate on the oblique asymptote. You know the y-value will reach infinity in the near of this asymptote. When you put this together with the roots of the polynomial you can go further with the form of the graph.
Both roots are smaller than the x-value of the oblique asymptote. Also the function goes for infinity for big negative x-values. So the graph will fall while you going to the first root but right after this it have to rise against since the graph has to reach the next root AND has to go for infinity at the oblique asymptote.
On the other hand the right side of the graph never hits the x-axis hence it goes from negative infinity up while the x-value is decreasing. So if the function would now go for infinity again there would be another interception with the x-axis. So it goes to negative infinity.
To put it all together: From the asymptotes you know how the function behaves at critical points such as infinity, negative infinity and at poles. Now you just focus on the values of the roots and compare them with the poles. From there on you can sketch the graph in a way of thinking how it develops at several x-values.
answered Jul 21 at 15:30
mrtaurho
700219
edited Jul 21 at 16:41
answered Jul 21 at 15:30
mrtaurho
700219
answered Jul 21 at 15:30
mrtaurho
700219
answered Jul 21 at 15:30
mrtaurho
700219
700219
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
add a comment |Â
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
What you mean by "concentrate on the horizontal asymptote". There is none. You mean Oblique Asymptote ?
– Arnuld
Jul 21 at 16:33
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
Yes, you are right. I will just fix this.
– mrtaurho
Jul 21 at 16:40
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
The x-intercept is at (1,0) and (4,0). How am I suppose to guess (without plotting) that it will cross only (1,0) but not (4,0) ? 2nd, A graph can cross slant/oblique and horizontal asymptotes (sometimes more than once). How can I know without plotting points that graph of current equation will not cross the oblique asymptote ?
– Arnuld
Jul 23 at 8:05
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858558%2funderstanding-graph-of-this-rational-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Instead of "vertical asymptote $(3,0)$" I would write: "vertical asymptote $x=3$". So the asymptote is a line, not a point.
– GEdgar
Jul 21 at 15:27