Can any non-vertical line be translated such that it intersects a parabola exactly twice? [closed]
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Given the parabola $y=ax^2+bx+c$ and the line $y=mx+d$, where $(a,b,m)not=0$ and $(a,b,c,m,d)inmathbb R$, does there exist an $m$ where, for all $d$, there is at most one solution to this system of equations? Or, in other words, does $$ax^2+bx+c=mx+dimplies ax^2+(b-m)x+(c-d)=0$$ have two solutions for some $d$?
algebra-precalculus analytic-geometry
edited Dec 25 '18 at 22:15
Namaste
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
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closed as off-topic by Namaste, Saad, Abcd, Cesareo, Davide Giraudo Dec 27 '18 at 22:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Saad, Abcd, Cesareo, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given the parabola $y=ax^2+bx+c$ and the line $y=mx+d$, where $(a,b,m)not=0$ and $(a,b,c,m,d)inmathbb R$, does there exist an $m$ where, for all $d$, there is at most one solution to this system of equations? Or, in other words, does $$ax^2+bx+c=mx+dimplies ax^2+(b-m)x+(c-d)=0$$ have two solutions for some $d$?
algebra-precalculus analytic-geometry
edited Dec 25 '18 at 22:15
Namaste
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
$endgroup$
closed as off-topic by Namaste, Saad, Abcd, Cesareo, Davide Giraudo Dec 27 '18 at 22:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Saad, Abcd, Cesareo, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
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Will the close voter please explain why this is "missing context or other details"?
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– DonielF
Dec 25 '18 at 22:26
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Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
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– Carl Schildkraut
Dec 26 '18 at 1:31
add a comment |
$begingroup$
Given the parabola $y=ax^2+bx+c$ and the line $y=mx+d$, where $(a,b,m)not=0$ and $(a,b,c,m,d)inmathbb R$, does there exist an $m$ where, for all $d$, there is at most one solution to this system of equations? Or, in other words, does $$ax^2+bx+c=mx+dimplies ax^2+(b-m)x+(c-d)=0$$ have two solutions for some $d$?
algebra-precalculus analytic-geometry
edited Dec 25 '18 at 22:15
Namaste
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
$endgroup$
Given the parabola $y=ax^2+bx+c$ and the line $y=mx+d$, where $(a,b,m)not=0$ and $(a,b,c,m,d)inmathbb R$, does there exist an $m$ where, for all $d$, there is at most one solution to this system of equations? Or, in other words, does $$ax^2+bx+c=mx+dimplies ax^2+(b-m)x+(c-d)=0$$ have two solutions for some $d$?
algebra-precalculus analytic-geometry
algebra-precalculus analytic-geometry
edited Dec 25 '18 at 22:15
Namaste
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
edited Dec 25 '18 at 22:15
Namaste
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
edited Dec 25 '18 at 22:15
Namaste
1
edited Dec 25 '18 at 22:15
Namaste
1
edited Dec 25 '18 at 22:15
Namaste
1
1
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
asked Dec 25 '18 at 22:09
DonielFDonielF
515515
515515
closed as off-topic by Namaste, Saad, Abcd, Cesareo, Davide Giraudo Dec 27 '18 at 22:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Saad, Abcd, Cesareo, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Saad, Abcd, Cesareo, Davide Giraudo Dec 27 '18 at 22:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Saad, Abcd, Cesareo, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
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Will the close voter please explain why this is "missing context or other details"?
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– DonielF
Dec 25 '18 at 22:26
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Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
$endgroup$
– Carl Schildkraut
Dec 26 '18 at 1:31
add a comment |
$begingroup$
Will the close voter please explain why this is "missing context or other details"?
$endgroup$
– DonielF
Dec 25 '18 at 22:26
$begingroup$
Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
$endgroup$
– Carl Schildkraut
Dec 26 '18 at 1:31
$begingroup$
Will the close voter please explain why this is "missing context or other details"?
$endgroup$
– DonielF
Dec 25 '18 at 22:26
$begingroup$
Will the close voter please explain why this is "missing context or other details"?
$endgroup$
– DonielF
Dec 25 '18 at 22:26
$begingroup$
Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
$endgroup$
– Carl Schildkraut
Dec 26 '18 at 1:31
$begingroup$
Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
$endgroup$
– Carl Schildkraut
Dec 26 '18 at 1:31
add a comment |
2 Answers
2
active
oldest
votes
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That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
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Duh. How did I miss this? Thanks!
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– DonielF
Dec 25 '18 at 22:27
add a comment |
$begingroup$
The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + text{sgn}(a).$
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
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How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
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– DonielF
Dec 25 '18 at 22:24
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@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
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– Display name
Dec 25 '18 at 22:26
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Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
add a comment |
$begingroup$
That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
add a comment |
$begingroup$
That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 25 '18 at 22:13
José Carlos SantosJosé Carlos Santos
173k23133242
173k23133242
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
add a comment |
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
$begingroup$
Duh. How did I miss this? Thanks!
$endgroup$
– DonielF
Dec 25 '18 at 22:27
add a comment |
$begingroup$
The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + text{sgn}(a).$
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
$endgroup$
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
add a comment |
$begingroup$
The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + text{sgn}(a).$
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
$endgroup$
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
add a comment |
$begingroup$
The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + text{sgn}(a).$
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
$endgroup$
The composition of 2 translations is a translation. First translate the parabola to the form $y = ax^2.$ Then you can translate the line to $y = mx + text{sgn}(a).$
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
answered Dec 25 '18 at 22:17
Display nameDisplay name
841314
841314
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
add a comment |
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
How does $y=ax^2+bx+c$ translate to $y=ax^2$? I can see how it translates to $y=ax^2+bx$, but how can you drop the $bx$ just by sliding it?
$endgroup$
– DonielF
Dec 25 '18 at 22:24
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
@DonielF Recall the vertex form: $ax^2 + bx + c = a(x-h)^2 + k.$ Now you can translate by $(h, -k).$
$endgroup$
– Display name
Dec 25 '18 at 22:26
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
$begingroup$
Vertex form, vertex form... I must have had a really bad algebra class if I never learned this. I can see how they're equivalent, and it's a really useful tool to have when dealing with these kinds of problems.
$endgroup$
– DonielF
Dec 25 '18 at 22:28
add a comment |
$begingroup$
Will the close voter please explain why this is "missing context or other details"?
$endgroup$
– DonielF
Dec 25 '18 at 22:26
$begingroup$
Can you please add what you've tried already and any particular difficulties you encountered? Maybe also add where you found this problem?
$endgroup$
– Carl Schildkraut
Dec 26 '18 at 1:31