Can any non-vertical line be translated such that it intersects a parabola exactly twice? [closed]












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$begingroup$


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







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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.
















  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$



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.
















  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Dec 25 '18 at 22:15









Namaste

1




1










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.












  • $begingroup$
    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










  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















3












$begingroup$

That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Duh. How did I miss this? Thanks!
    $endgroup$
    – DonielF
    Dec 25 '18 at 22:27



















2












$begingroup$

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 $y = mx + text{sgn}(a).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Duh. How did I miss this? Thanks!
    $endgroup$
    – DonielF
    Dec 25 '18 at 22:27
















3












$begingroup$

That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Duh. How did I miss this? Thanks!
    $endgroup$
    – DonielF
    Dec 25 '18 at 22:27














3












3








3





$begingroup$

That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?






share|cite|improve this answer









$endgroup$



That equation will have two solutions if and only if$$(b-m)^2-4a(c-d)>0.$$Can you take it from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











2












$begingroup$

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 $y = mx + text{sgn}(a).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $y = mx + text{sgn}(a).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $y = mx + text{sgn}(a).$






share|cite|improve this answer









$endgroup$



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 $y = mx + text{sgn}(a).$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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