How can I find the equation of one parabola given the equation of another parabola?
The equation of the other parabola has to follow the form:
$4p(x – h) = (y – k)^2$
because it is a sideways parabola. I can see that the vertex is at (-4.5,18)
So then the equation would be $4p(x+4.5) = (y-18)^2$
Now I can just plug in a value like (4,5) for x and y in that equation and solve for p.
If I couldn't see the values of the vertex on the graph, how would I solve this problem?
quadratics
asked Nov 24 at 18:23
user477465
1596
add a comment |
The equation of the other parabola has to follow the form:
$4p(x – h) = (y – k)^2$
because it is a sideways parabola. I can see that the vertex is at (-4.5,18)
So then the equation would be $4p(x+4.5) = (y-18)^2$
Now I can just plug in a value like (4,5) for x and y in that equation and solve for p.
If I couldn't see the values of the vertex on the graph, how would I solve this problem?
quadratics
asked Nov 24 at 18:23
user477465
1596
add a comment |
The equation of the other parabola has to follow the form:
$4p(x – h) = (y – k)^2$
because it is a sideways parabola. I can see that the vertex is at (-4.5,18)
So then the equation would be $4p(x+4.5) = (y-18)^2$
Now I can just plug in a value like (4,5) for x and y in that equation and solve for p.
If I couldn't see the values of the vertex on the graph, how would I solve this problem?
quadratics
asked Nov 24 at 18:23
user477465
1596
The equation of the other parabola has to follow the form:
$4p(x – h) = (y – k)^2$
because it is a sideways parabola. I can see that the vertex is at (-4.5,18)
So then the equation would be $4p(x+4.5) = (y-18)^2$
Now I can just plug in a value like (4,5) for x and y in that equation and solve for p.
If I couldn't see the values of the vertex on the graph, how would I solve this problem?
quadratics
quadratics
asked Nov 24 at 18:23
user477465
1596
asked Nov 24 at 18:23
user477465
1596
asked Nov 24 at 18:23
user477465
1596
asked Nov 24 at 18:23
user477465
1596
asked Nov 24 at 18:23
user477465
1596
1596
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Assuming the axis of the solid parabola is parallel to the $x$-axis...
Plug in the points. begin{align*}
4 p(4 - h) &= (5 - k)^2 text{,} \
4 p(-2 - h) &= (11 - k)^2 text{, and} \
4 p(-4 - h) &= (21 - k)^2 text{.}
end{align*}
Then solve for $h$, $p$, and $k$. You'll discover $h neq -4.5$.
(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)
answered Nov 24 at 18:33
Eric Towers
31.8k22265
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
add a comment |
The best solution is to write three equations using the three given points:
Eq. 1: $4p(4-h)=(5−k)^2$
Eq. 2: $4p(-2-h)=(11−k)^2$
Eq. 3: $4p(-4-h)=(21−k)^2$
And use substitution to solve for p, k, and h. To begin:
From Eq. 1: $h=4-(5-k)^2/4p$
Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$
Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$
Simplify: $-24p+25=121-12k$
Simplify: $p=0.5k-4$
From there, it should be relatively simple to find $h$ and $k$.
answered Nov 24 at 18:32
chokfull
112
add a comment |
There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.
In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).
By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)
answered Nov 24 at 18:30
TonyK
41.4k353132
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
add a comment |
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3 Answers
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3 Answers
3
active
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Assuming the axis of the solid parabola is parallel to the $x$-axis...
Plug in the points. begin{align*}
4 p(4 - h) &= (5 - k)^2 text{,} \
4 p(-2 - h) &= (11 - k)^2 text{, and} \
4 p(-4 - h) &= (21 - k)^2 text{.}
end{align*}
Then solve for $h$, $p$, and $k$. You'll discover $h neq -4.5$.
(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)
answered Nov 24 at 18:33
Eric Towers
31.8k22265
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
add a comment |
Assuming the axis of the solid parabola is parallel to the $x$-axis...
Plug in the points. begin{align*}
4 p(4 - h) &= (5 - k)^2 text{,} \
4 p(-2 - h) &= (11 - k)^2 text{, and} \
4 p(-4 - h) &= (21 - k)^2 text{.}
end{align*}
Then solve for $h$, $p$, and $k$. You'll discover $h neq -4.5$.
(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)
answered Nov 24 at 18:33
Eric Towers
31.8k22265
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
add a comment |
Assuming the axis of the solid parabola is parallel to the $x$-axis...
Plug in the points. begin{align*}
4 p(4 - h) &= (5 - k)^2 text{,} \
4 p(-2 - h) &= (11 - k)^2 text{, and} \
4 p(-4 - h) &= (21 - k)^2 text{.}
end{align*}
Then solve for $h$, $p$, and $k$. You'll discover $h neq -4.5$.
(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)
answered Nov 24 at 18:33
Eric Towers
31.8k22265
Assuming the axis of the solid parabola is parallel to the $x$-axis...
Plug in the points. begin{align*}
4 p(4 - h) &= (5 - k)^2 text{,} \
4 p(-2 - h) &= (11 - k)^2 text{, and} \
4 p(-4 - h) &= (21 - k)^2 text{.}
end{align*}
Then solve for $h$, $p$, and $k$. You'll discover $h neq -4.5$.
(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)
answered Nov 24 at 18:33
Eric Towers
31.8k22265
answered Nov 24 at 18:33
Eric Towers
31.8k22265
answered Nov 24 at 18:33
Eric Towers
31.8k22265
answered Nov 24 at 18:33
Eric Towers
31.8k22265
31.8k22265
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
add a comment |
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
when i solve this, i get $k=28$ and $p = 10$ and when i plug back in, i get $h = -9.225$ These values don't make any sense. If you look at the picture, the vertex isn't at $(-9.225,28)$. Did i get my values wrong?
– user477465
Nov 25 at 7:20
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
also, what does the first (dashed) parabola have to do with solving for the second? they don't seem related
– user477465
Nov 25 at 7:21
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : When I solve for $p$ and $k$, I get $k = 18$. Since the diagram shows $11 < k < 21$, this seems plausible.
– Eric Towers
Nov 25 at 16:27
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
@user477465 : The dashed parabola reminds the student that solving for $y$ as a function of $x$ will produce a parabola, but it is not the desired parabola.
– Eric Towers
Nov 25 at 16:32
add a comment |
The best solution is to write three equations using the three given points:
Eq. 1: $4p(4-h)=(5−k)^2$
Eq. 2: $4p(-2-h)=(11−k)^2$
Eq. 3: $4p(-4-h)=(21−k)^2$
And use substitution to solve for p, k, and h. To begin:
From Eq. 1: $h=4-(5-k)^2/4p$
Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$
Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$
Simplify: $-24p+25=121-12k$
Simplify: $p=0.5k-4$
From there, it should be relatively simple to find $h$ and $k$.
answered Nov 24 at 18:32
chokfull
112
add a comment |
The best solution is to write three equations using the three given points:
Eq. 1: $4p(4-h)=(5−k)^2$
Eq. 2: $4p(-2-h)=(11−k)^2$
Eq. 3: $4p(-4-h)=(21−k)^2$
And use substitution to solve for p, k, and h. To begin:
From Eq. 1: $h=4-(5-k)^2/4p$
Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$
Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$
Simplify: $-24p+25=121-12k$
Simplify: $p=0.5k-4$
From there, it should be relatively simple to find $h$ and $k$.
answered Nov 24 at 18:32
chokfull
112
add a comment |
The best solution is to write three equations using the three given points:
Eq. 1: $4p(4-h)=(5−k)^2$
Eq. 2: $4p(-2-h)=(11−k)^2$
Eq. 3: $4p(-4-h)=(21−k)^2$
And use substitution to solve for p, k, and h. To begin:
From Eq. 1: $h=4-(5-k)^2/4p$
Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$
Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$
Simplify: $-24p+25=121-12k$
Simplify: $p=0.5k-4$
From there, it should be relatively simple to find $h$ and $k$.
answered Nov 24 at 18:32
chokfull
112
The best solution is to write three equations using the three given points:
Eq. 1: $4p(4-h)=(5−k)^2$
Eq. 2: $4p(-2-h)=(11−k)^2$
Eq. 3: $4p(-4-h)=(21−k)^2$
And use substitution to solve for p, k, and h. To begin:
From Eq. 1: $h=4-(5-k)^2/4p$
Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$
Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$
Simplify: $-24p+25=121-12k$
Simplify: $p=0.5k-4$
From there, it should be relatively simple to find $h$ and $k$.
answered Nov 24 at 18:32
chokfull
112
edited Nov 24 at 18:56
answered Nov 24 at 18:32
chokfull
112
answered Nov 24 at 18:32
chokfull
112
answered Nov 24 at 18:32
chokfull
112
112
add a comment |
add a comment |
There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.
In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).
By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)
answered Nov 24 at 18:30
TonyK
41.4k353132
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
add a comment |
There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.
In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).
By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)
answered Nov 24 at 18:30
TonyK
41.4k353132
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
add a comment |
There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.
In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).
By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)
answered Nov 24 at 18:30
TonyK
41.4k353132
There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.
In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).
By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)
answered Nov 24 at 18:30
TonyK
41.4k353132
edited Nov 24 at 18:38
answered Nov 24 at 18:30
TonyK
41.4k353132
answered Nov 24 at 18:30
TonyK
41.4k353132
answered Nov 24 at 18:30
TonyK
41.4k353132
41.4k353132
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
add a comment |
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
I was just looking at the axes and seeing that it lined up with -4.5 and 18 but you're right, that isn't correct.
– user477465
Nov 24 at 18:39
add a comment |
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