How to sketch an equation with variables x and y












0












$begingroup$


I was doing a sample exam for my final tommorow and I did not understand how to solve the following question. Maybe it's relevant to note that during this math course we have been talking about the Implicit Function Theorem and Inverse Mapping Theorem.



We were supposed to sketch the equation g(x,y), but I have no clue how they managed to come to the conclusion of that in the picture. For example, how do they know its a hyperbola and how did they find those asymptotes in a multi-variate equation? I would greatly appreciate any input.



Thank you very much in advance.



enter image description here






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  • $begingroup$
    set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
    $endgroup$
    – Doug M
    Dec 20 '18 at 18:57
















0












$begingroup$


I was doing a sample exam for my final tommorow and I did not understand how to solve the following question. Maybe it's relevant to note that during this math course we have been talking about the Implicit Function Theorem and Inverse Mapping Theorem.



We were supposed to sketch the equation g(x,y), but I have no clue how they managed to come to the conclusion of that in the picture. For example, how do they know its a hyperbola and how did they find those asymptotes in a multi-variate equation? I would greatly appreciate any input.



Thank you very much in advance.



enter image description here






implicit-function-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
    $endgroup$
    – Doug M
    Dec 20 '18 at 18:57














0












0








0





$begingroup$


I was doing a sample exam for my final tommorow and I did not understand how to solve the following question. Maybe it's relevant to note that during this math course we have been talking about the Implicit Function Theorem and Inverse Mapping Theorem.



We were supposed to sketch the equation g(x,y), but I have no clue how they managed to come to the conclusion of that in the picture. For example, how do they know its a hyperbola and how did they find those asymptotes in a multi-variate equation? I would greatly appreciate any input.



Thank you very much in advance.



enter image description here






implicit-function-theorem







share|cite|improve this question









$endgroup$




I was doing a sample exam for my final tommorow and I did not understand how to solve the following question. Maybe it's relevant to note that during this math course we have been talking about the Implicit Function Theorem and Inverse Mapping Theorem.



We were supposed to sketch the equation g(x,y), but I have no clue how they managed to come to the conclusion of that in the picture. For example, how do they know its a hyperbola and how did they find those asymptotes in a multi-variate equation? I would greatly appreciate any input.



Thank you very much in advance.



enter image description here






implicit-function-theorem




implicit-function-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 18:51









MathNoob123MathNoob123

624




624












  • $begingroup$
    set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
    $endgroup$
    – Doug M
    Dec 20 '18 at 18:57


















  • $begingroup$
    set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
    $endgroup$
    – Doug M
    Dec 20 '18 at 18:57
















$begingroup$
set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
$endgroup$
– Doug M
Dec 20 '18 at 18:57




$begingroup$
set $g(x,y) = 0$ and you have the equation for a hyperbola, that you learned how to graph in algebra class.
$endgroup$
– Doug M
Dec 20 '18 at 18:57










1 Answer
1






active

oldest

votes


















1












$begingroup$

Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = {(x,y) in mathbb R^2 : g(x,y) = 0}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = {(x,y) in mathbb R^2 : z = 0},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.



Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$begin{align}x^2 - y^2 + 2y - 2 & = 0 \ x^2 -1(y^2 - 2y) - 2 &= 0 tag{Factor out $-1$ from $y$'s}\ x^2 - (underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 tag{Complete square}\ x^2 - [(y-1)^2 - 1] - 2 &= 0\ x^2 - (y-1)^2 + 1 - 2 &= 0 tag{Simplify}\ x^2 - (y-1)^2 - 1 &= 0 tag{Simplify}\ (x - 0)^2 - (y - 1)^2 = 1. tag{Move $-1$ over}end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much, I greatly appreciate the help. I completely understand it now!
    $endgroup$
    – MathNoob123
    Dec 20 '18 at 21:42












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1 Answer
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1 Answer
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1












$begingroup$

Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = {(x,y) in mathbb R^2 : g(x,y) = 0}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = {(x,y) in mathbb R^2 : z = 0},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.



Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$begin{align}x^2 - y^2 + 2y - 2 & = 0 \ x^2 -1(y^2 - 2y) - 2 &= 0 tag{Factor out $-1$ from $y$'s}\ x^2 - (underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 tag{Complete square}\ x^2 - [(y-1)^2 - 1] - 2 &= 0\ x^2 - (y-1)^2 + 1 - 2 &= 0 tag{Simplify}\ x^2 - (y-1)^2 - 1 &= 0 tag{Simplify}\ (x - 0)^2 - (y - 1)^2 = 1. tag{Move $-1$ over}end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much, I greatly appreciate the help. I completely understand it now!
    $endgroup$
    – MathNoob123
    Dec 20 '18 at 21:42
















1












$begingroup$

Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = {(x,y) in mathbb R^2 : g(x,y) = 0}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = {(x,y) in mathbb R^2 : z = 0},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.



Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$begin{align}x^2 - y^2 + 2y - 2 & = 0 \ x^2 -1(y^2 - 2y) - 2 &= 0 tag{Factor out $-1$ from $y$'s}\ x^2 - (underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 tag{Complete square}\ x^2 - [(y-1)^2 - 1] - 2 &= 0\ x^2 - (y-1)^2 + 1 - 2 &= 0 tag{Simplify}\ x^2 - (y-1)^2 - 1 &= 0 tag{Simplify}\ (x - 0)^2 - (y - 1)^2 = 1. tag{Move $-1$ over}end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much, I greatly appreciate the help. I completely understand it now!
    $endgroup$
    – MathNoob123
    Dec 20 '18 at 21:42














1












1








1





$begingroup$

Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = {(x,y) in mathbb R^2 : g(x,y) = 0}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = {(x,y) in mathbb R^2 : z = 0},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.



Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$begin{align}x^2 - y^2 + 2y - 2 & = 0 \ x^2 -1(y^2 - 2y) - 2 &= 0 tag{Factor out $-1$ from $y$'s}\ x^2 - (underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 tag{Complete square}\ x^2 - [(y-1)^2 - 1] - 2 &= 0\ x^2 - (y-1)^2 + 1 - 2 &= 0 tag{Simplify}\ x^2 - (y-1)^2 - 1 &= 0 tag{Simplify}\ (x - 0)^2 - (y - 1)^2 = 1. tag{Move $-1$ over}end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.






share|cite|improve this answer











$endgroup$



Upon initial inspection, there are some key points the problem asks you to consider. For example, notice that $V = {(x,y) in mathbb R^2 : g(x,y) = 0}$ is the set of all points on the $xy$-plane where $g$ is $0$. To see this, note that $g(x,y)$ describes a surface in $3D$ (three-dimensions) given that it can be described by two variables. Hence, let $z := g(x,y) = x^2 - y^2 + 2y - 2$. Then $V$ is written $$V = {(x,y) in mathbb R^2 : z = 0},$$ and in $3D$, taking $z=0$ means the $xy$-plane itself. This is why the writers set the function $g$ equal to $0$.



Next, they attempt to complete the square for both $x$ and $y$ (and really, only $y$ will require any work here) in $g$ so that it is more immediate what the function is: $$begin{align}x^2 - y^2 + 2y - 2 & = 0 \ x^2 -1(y^2 - 2y) - 2 &= 0 tag{Factor out $-1$ from $y$'s}\ x^2 - (underbrace{y^2 - 2y + 1}_{(y-1)^2} - 1) - 2 &= 0 tag{Complete square}\ x^2 - [(y-1)^2 - 1] - 2 &= 0\ x^2 - (y-1)^2 + 1 - 2 &= 0 tag{Simplify}\ x^2 - (y-1)^2 - 1 &= 0 tag{Simplify}\ (x - 0)^2 - (y - 1)^2 = 1. tag{Move $-1$ over}end{align}$$ Clearly, the end result is the standard form of a hyperbola in $2D$, which is how the authors make this deduction. They then proceed to plot the hyperbola as described.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 4:56

























answered Dec 20 '18 at 19:03









Decaf-MathDecaf-Math

3,4221026




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  • $begingroup$
    Thank you very much, I greatly appreciate the help. I completely understand it now!
    $endgroup$
    – MathNoob123
    Dec 20 '18 at 21:42


















  • $begingroup$
    Thank you very much, I greatly appreciate the help. I completely understand it now!
    $endgroup$
    – MathNoob123
    Dec 20 '18 at 21:42
















$begingroup$
Thank you very much, I greatly appreciate the help. I completely understand it now!
$endgroup$
– MathNoob123
Dec 20 '18 at 21:42




$begingroup$
Thank you very much, I greatly appreciate the help. I completely understand it now!
$endgroup$
– MathNoob123
Dec 20 '18 at 21:42


















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