Mixing
How to transform Parabola to 3D graph and lift the vertex upon a vertical line (z axis)?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
If I have a quadratic in the form of $f(x) =ax^2+bx+c$, how would I be able to transfer this onto a 3D Graph where $x$=horizontal, $y$=the other 2D dimension and $z$= height , and raise the turning point by $k$ in the $z$ axis, how might I find the equation of this 3D Graph$?$
Diagram (1) visualisation
You could imagine the vertex of the green curve (the parabola), being shifted up/down the red vertical line by $k$ units, while the $x$ intercepts stay still. And then that is the new function I need the equation for.
Diagram (2) angle calculations
In that picture, I have made the quadratic: $f(x)=-x^2+10x$, and I want to shift it up by $1$ unit in the $z$ axis. I have drawn what that new function would look like with a red pen.
I am keeping the $x$ intercepts the same, except since I am pulling the vertex upwards, the distance $AB>AC$, (look carefully at the diagram to see the letters A B and C on the graph), so the quadratic would have stretched upwards. You could imagine that $BC$ is part of the line that I am shifting the vertex upon.
I let $k=theta=angle BAC$,
and draw a triangle $triangle ABC$ onto the 3D graph, where $A$ is at the $x$ coordinate of the turning point of $f(x)$ which is $x=-b/2a=-10/-2=5$
Only the $x$ value though so I put it on the $x$ axis at $x=5to (5,0,0)$
I make point $B$ at the turning point of my lifted curve $(x,y,z)to(5,25,1)$
I make the point $C$ at the vertex of the original function $f(x)$ which is at $(5,25,0)$.
It is quite simple now to draw the triangle seperate from the graph. All lengths from here on are absolute and don't factor in angles.
Distance $AC$ = $25$ units, $BC$=$1$ unit, and $AB = sqrtAB^2+BC^2 = sqrt626$ units. And by using basic trigonometry, $theta=angle BAC approx 2.29°$
How do I get my new function?
multivariable-calculus quadratics transformation 3d
asked Jul 28 at 8:07
Simplex1
82
add a comment |Â
up vote
0
down vote
favorite
If I have a quadratic in the form of $f(x) =ax^2+bx+c$, how would I be able to transfer this onto a 3D Graph where $x$=horizontal, $y$=the other 2D dimension and $z$= height , and raise the turning point by $k$ in the $z$ axis, how might I find the equation of this 3D Graph$?$
Diagram (1) visualisation
You could imagine the vertex of the green curve (the parabola), being shifted up/down the red vertical line by $k$ units, while the $x$ intercepts stay still. And then that is the new function I need the equation for.
Diagram (2) angle calculations
In that picture, I have made the quadratic: $f(x)=-x^2+10x$, and I want to shift it up by $1$ unit in the $z$ axis. I have drawn what that new function would look like with a red pen.
I am keeping the $x$ intercepts the same, except since I am pulling the vertex upwards, the distance $AB>AC$, (look carefully at the diagram to see the letters A B and C on the graph), so the quadratic would have stretched upwards. You could imagine that $BC$ is part of the line that I am shifting the vertex upon.
I let $k=theta=angle BAC$,
and draw a triangle $triangle ABC$ onto the 3D graph, where $A$ is at the $x$ coordinate of the turning point of $f(x)$ which is $x=-b/2a=-10/-2=5$
Only the $x$ value though so I put it on the $x$ axis at $x=5to (5,0,0)$
I make point $B$ at the turning point of my lifted curve $(x,y,z)to(5,25,1)$
I make the point $C$ at the vertex of the original function $f(x)$ which is at $(5,25,0)$.
It is quite simple now to draw the triangle seperate from the graph. All lengths from here on are absolute and don't factor in angles.
Distance $AC$ = $25$ units, $BC$=$1$ unit, and $AB = sqrtAB^2+BC^2 = sqrt626$ units. And by using basic trigonometry, $theta=angle BAC approx 2.29°$
How do I get my new function?
multivariable-calculus quadratics transformation 3d
asked Jul 28 at 8:07
Simplex1
82
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I have a quadratic in the form of $f(x) =ax^2+bx+c$, how would I be able to transfer this onto a 3D Graph where $x$=horizontal, $y$=the other 2D dimension and $z$= height , and raise the turning point by $k$ in the $z$ axis, how might I find the equation of this 3D Graph$?$
Diagram (1) visualisation
You could imagine the vertex of the green curve (the parabola), being shifted up/down the red vertical line by $k$ units, while the $x$ intercepts stay still. And then that is the new function I need the equation for.
Diagram (2) angle calculations
In that picture, I have made the quadratic: $f(x)=-x^2+10x$, and I want to shift it up by $1$ unit in the $z$ axis. I have drawn what that new function would look like with a red pen.
I am keeping the $x$ intercepts the same, except since I am pulling the vertex upwards, the distance $AB>AC$, (look carefully at the diagram to see the letters A B and C on the graph), so the quadratic would have stretched upwards. You could imagine that $BC$ is part of the line that I am shifting the vertex upon.
I let $k=theta=angle BAC$,
and draw a triangle $triangle ABC$ onto the 3D graph, where $A$ is at the $x$ coordinate of the turning point of $f(x)$ which is $x=-b/2a=-10/-2=5$
Only the $x$ value though so I put it on the $x$ axis at $x=5to (5,0,0)$
I make point $B$ at the turning point of my lifted curve $(x,y,z)to(5,25,1)$
I make the point $C$ at the vertex of the original function $f(x)$ which is at $(5,25,0)$.
It is quite simple now to draw the triangle seperate from the graph. All lengths from here on are absolute and don't factor in angles.
Distance $AC$ = $25$ units, $BC$=$1$ unit, and $AB = sqrtAB^2+BC^2 = sqrt626$ units. And by using basic trigonometry, $theta=angle BAC approx 2.29°$
How do I get my new function?
multivariable-calculus quadratics transformation 3d
asked Jul 28 at 8:07
Simplex1
82
If I have a quadratic in the form of $f(x) =ax^2+bx+c$, how would I be able to transfer this onto a 3D Graph where $x$=horizontal, $y$=the other 2D dimension and $z$= height , and raise the turning point by $k$ in the $z$ axis, how might I find the equation of this 3D Graph$?$
Diagram (1) visualisation
You could imagine the vertex of the green curve (the parabola), being shifted up/down the red vertical line by $k$ units, while the $x$ intercepts stay still. And then that is the new function I need the equation for.
Diagram (2) angle calculations
In that picture, I have made the quadratic: $f(x)=-x^2+10x$, and I want to shift it up by $1$ unit in the $z$ axis. I have drawn what that new function would look like with a red pen.
I am keeping the $x$ intercepts the same, except since I am pulling the vertex upwards, the distance $AB>AC$, (look carefully at the diagram to see the letters A B and C on the graph), so the quadratic would have stretched upwards. You could imagine that $BC$ is part of the line that I am shifting the vertex upon.
I let $k=theta=angle BAC$,
and draw a triangle $triangle ABC$ onto the 3D graph, where $A$ is at the $x$ coordinate of the turning point of $f(x)$ which is $x=-b/2a=-10/-2=5$
Only the $x$ value though so I put it on the $x$ axis at $x=5to (5,0,0)$
I make point $B$ at the turning point of my lifted curve $(x,y,z)to(5,25,1)$
I make the point $C$ at the vertex of the original function $f(x)$ which is at $(5,25,0)$.
It is quite simple now to draw the triangle seperate from the graph. All lengths from here on are absolute and don't factor in angles.
Distance $AC$ = $25$ units, $BC$=$1$ unit, and $AB = sqrtAB^2+BC^2 = sqrt626$ units. And by using basic trigonometry, $theta=angle BAC approx 2.29°$
How do I get my new function?
multivariable-calculus quadratics transformation 3d
asked Jul 28 at 8:07
Simplex1
82
edited Jul 28 at 8:13
asked Jul 28 at 8:07
Simplex1
82
asked Jul 28 at 8:07
Simplex1
82
asked Jul 28 at 8:07
Simplex1
82
82
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27
add a comment |Â
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
So the equation of your curve, before transformation, is this:
$$
y = ax^2 + bx + c \
z = 0
$$
And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form:
$$
y = (x-u)^2 + w\
z = 0
$$
which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w ne 0$. Then the intercepts occur when $y = 0$, i.e., when
beginalign
(x-u)^2 +w &= 0 \
(x-u)^2 &= -w \
(x-u) &= pmsqrt-w \
x &= u pm sqrt-w
endalign
So in the case where $w < 0$, this has solutions; for $w ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at
$$
(u - sqrt-w, 0, 0)\
(u + sqrt-w, 0, 0)
$$
and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map
$$
(x, y, z) mapsto (x, y, z + px + qy)
$$
where the $y = 0$ line remains fixed, so
$$
(x, 0, z) mapsto (x, 0, z + px) = (x, 0, z)
$$
hence $p = 0$.
And we also want to have
$$
(u, w, 0) mapsto (u, w, H)
$$
for some target height $H$. So we need
$$
(u, w, 0) mapsto (u, w, 0 + qw) = (u, w, H)
$$
hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map
$$
(x, y, z) mapsto (x, y, z + fracHw y)
$$
The "before" equation was
beginalign
y &= (x-u)^2 + w\
z &= 0
endalign
The "after" equation is therefore
beginalign
y &= (x-u)^2 + w\
z &= fracHwleft((x-u)^2 + w right)
endalign
Let's make this concrete with an example. Suppose we have
$$
y = x^2 + 10x
$$
which we rewrite in vertex form as
$$
y = (x+5)^2 - 15
$$
so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot
beginalign
y &= (x+5)^2 - 15\
z &= frac3-15left( (x+5)^2 - 15 right)
endalign
Here's the plotted result, using geogebra:
answered Jul 28 at 9:47
John Hughes
59.4k23785
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
So the equation of your curve, before transformation, is this:
$$
y = ax^2 + bx + c \
z = 0
$$
And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form:
$$
y = (x-u)^2 + w\
z = 0
$$
which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w ne 0$. Then the intercepts occur when $y = 0$, i.e., when
beginalign
(x-u)^2 +w &= 0 \
(x-u)^2 &= -w \
(x-u) &= pmsqrt-w \
x &= u pm sqrt-w
endalign
So in the case where $w < 0$, this has solutions; for $w ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at
$$
(u - sqrt-w, 0, 0)\
(u + sqrt-w, 0, 0)
$$
and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map
$$
(x, y, z) mapsto (x, y, z + px + qy)
$$
where the $y = 0$ line remains fixed, so
$$
(x, 0, z) mapsto (x, 0, z + px) = (x, 0, z)
$$
hence $p = 0$.
And we also want to have
$$
(u, w, 0) mapsto (u, w, H)
$$
for some target height $H$. So we need
$$
(u, w, 0) mapsto (u, w, 0 + qw) = (u, w, H)
$$
hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map
$$
(x, y, z) mapsto (x, y, z + fracHw y)
$$
The "before" equation was
beginalign
y &= (x-u)^2 + w\
z &= 0
endalign
The "after" equation is therefore
beginalign
y &= (x-u)^2 + w\
z &= fracHwleft((x-u)^2 + w right)
endalign
Let's make this concrete with an example. Suppose we have
$$
y = x^2 + 10x
$$
which we rewrite in vertex form as
$$
y = (x+5)^2 - 15
$$
so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot
beginalign
y &= (x+5)^2 - 15\
z &= frac3-15left( (x+5)^2 - 15 right)
endalign
Here's the plotted result, using geogebra:
answered Jul 28 at 9:47
John Hughes
59.4k23785
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
add a comment |Â
up vote
0
down vote
accepted
So the equation of your curve, before transformation, is this:
$$
y = ax^2 + bx + c \
z = 0
$$
And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form:
$$
y = (x-u)^2 + w\
z = 0
$$
which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w ne 0$. Then the intercepts occur when $y = 0$, i.e., when
beginalign
(x-u)^2 +w &= 0 \
(x-u)^2 &= -w \
(x-u) &= pmsqrt-w \
x &= u pm sqrt-w
endalign
So in the case where $w < 0$, this has solutions; for $w ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at
$$
(u - sqrt-w, 0, 0)\
(u + sqrt-w, 0, 0)
$$
and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map
$$
(x, y, z) mapsto (x, y, z + px + qy)
$$
where the $y = 0$ line remains fixed, so
$$
(x, 0, z) mapsto (x, 0, z + px) = (x, 0, z)
$$
hence $p = 0$.
And we also want to have
$$
(u, w, 0) mapsto (u, w, H)
$$
for some target height $H$. So we need
$$
(u, w, 0) mapsto (u, w, 0 + qw) = (u, w, H)
$$
hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map
$$
(x, y, z) mapsto (x, y, z + fracHw y)
$$
The "before" equation was
beginalign
y &= (x-u)^2 + w\
z &= 0
endalign
The "after" equation is therefore
beginalign
y &= (x-u)^2 + w\
z &= fracHwleft((x-u)^2 + w right)
endalign
Let's make this concrete with an example. Suppose we have
$$
y = x^2 + 10x
$$
which we rewrite in vertex form as
$$
y = (x+5)^2 - 15
$$
so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot
beginalign
y &= (x+5)^2 - 15\
z &= frac3-15left( (x+5)^2 - 15 right)
endalign
Here's the plotted result, using geogebra:
answered Jul 28 at 9:47
John Hughes
59.4k23785
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
So the equation of your curve, before transformation, is this:
$$
y = ax^2 + bx + c \
z = 0
$$
And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form:
$$
y = (x-u)^2 + w\
z = 0
$$
which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w ne 0$. Then the intercepts occur when $y = 0$, i.e., when
beginalign
(x-u)^2 +w &= 0 \
(x-u)^2 &= -w \
(x-u) &= pmsqrt-w \
x &= u pm sqrt-w
endalign
So in the case where $w < 0$, this has solutions; for $w ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at
$$
(u - sqrt-w, 0, 0)\
(u + sqrt-w, 0, 0)
$$
and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map
$$
(x, y, z) mapsto (x, y, z + px + qy)
$$
where the $y = 0$ line remains fixed, so
$$
(x, 0, z) mapsto (x, 0, z + px) = (x, 0, z)
$$
hence $p = 0$.
And we also want to have
$$
(u, w, 0) mapsto (u, w, H)
$$
for some target height $H$. So we need
$$
(u, w, 0) mapsto (u, w, 0 + qw) = (u, w, H)
$$
hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map
$$
(x, y, z) mapsto (x, y, z + fracHw y)
$$
The "before" equation was
beginalign
y &= (x-u)^2 + w\
z &= 0
endalign
The "after" equation is therefore
beginalign
y &= (x-u)^2 + w\
z &= fracHwleft((x-u)^2 + w right)
endalign
Let's make this concrete with an example. Suppose we have
$$
y = x^2 + 10x
$$
which we rewrite in vertex form as
$$
y = (x+5)^2 - 15
$$
so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot
beginalign
y &= (x+5)^2 - 15\
z &= frac3-15left( (x+5)^2 - 15 right)
endalign
Here's the plotted result, using geogebra:
answered Jul 28 at 9:47
John Hughes
59.4k23785
So the equation of your curve, before transformation, is this:
$$
y = ax^2 + bx + c \
z = 0
$$
And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form:
$$
y = (x-u)^2 + w\
z = 0
$$
which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w ne 0$. Then the intercepts occur when $y = 0$, i.e., when
beginalign
(x-u)^2 +w &= 0 \
(x-u)^2 &= -w \
(x-u) &= pmsqrt-w \
x &= u pm sqrt-w
endalign
So in the case where $w < 0$, this has solutions; for $w ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at
$$
(u - sqrt-w, 0, 0)\
(u + sqrt-w, 0, 0)
$$
and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map
$$
(x, y, z) mapsto (x, y, z + px + qy)
$$
where the $y = 0$ line remains fixed, so
$$
(x, 0, z) mapsto (x, 0, z + px) = (x, 0, z)
$$
hence $p = 0$.
And we also want to have
$$
(u, w, 0) mapsto (u, w, H)
$$
for some target height $H$. So we need
$$
(u, w, 0) mapsto (u, w, 0 + qw) = (u, w, H)
$$
hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map
$$
(x, y, z) mapsto (x, y, z + fracHw y)
$$
The "before" equation was
beginalign
y &= (x-u)^2 + w\
z &= 0
endalign
The "after" equation is therefore
beginalign
y &= (x-u)^2 + w\
z &= fracHwleft((x-u)^2 + w right)
endalign
Let's make this concrete with an example. Suppose we have
$$
y = x^2 + 10x
$$
which we rewrite in vertex form as
$$
y = (x+5)^2 - 15
$$
so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot
beginalign
y &= (x+5)^2 - 15\
z &= frac3-15left( (x+5)^2 - 15 right)
endalign
Here's the plotted result, using geogebra:
answered Jul 28 at 9:47
John Hughes
59.4k23785
edited Jul 28 at 10:11
answered Jul 28 at 9:47
John Hughes
59.4k23785
answered Jul 28 at 9:47
John Hughes
59.4k23785
answered Jul 28 at 9:47
John Hughes
59.4k23785
59.4k23785
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
add a comment |Â
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
If the equation in vertex form has a coefficient, like $a(x-u)^2+w$, would the result be be $z=(H/w)(a(x-u)^2+w)$?
– Simplex1
Jul 28 at 10:53
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Yes; the only thing used in computing the new $z$ formula was $H/w$, where $H$ is the desired height, and $(u, w)$ is the vertex. Since multiplying $(x-u)^2$ by a constant doesn't change the location of its min ($x = u$, of course), everything still works.
– John Hughes
Jul 28 at 10:55
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
Thank you for the clear answer to this problem.
– Simplex1
Jul 28 at 11:16
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%2f2865090%2fhow-to-transform-parabola-to-3d-graph-and-lift-the-vertex-upon-a-vertical-line%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
Try the intersection between $S(x,y,z) = y-(a x^2+ b x + c)=0$ and $Pito alpha x+beta y+gamma z+delta = 0$ giving $-a beta ^2 y^2-2 a beta delta y-2 a beta gamma y z-a delta ^2-2 a delta gamma z-a gamma ^2 z^2-alpha ^2 c+alpha ^2 y+alpha b beta y+alpha b delta +alpha b gamma z=0$
– Cesareo
Jul 28 at 9:27