finding a tangent line to a parabola
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I am practicing for a math contest and I encountered the following problem that I don't know how to solve:
For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?
How do I solve it?
algebra-precalculus contest-math
New contributor
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show 3 more comments
$begingroup$
I am practicing for a math contest and I encountered the following problem that I don't know how to solve:
For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?
How do I solve it?
algebra-precalculus contest-math
New contributor
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1
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Do you know how to find a derivative?
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– R. Burton
5 hours ago
1
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Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
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– the_fox
5 hours ago
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A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
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– NoChance
4 hours ago
1
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@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
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– amd
3 hours ago
1
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@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago
|
show 3 more comments
$begingroup$
I am practicing for a math contest and I encountered the following problem that I don't know how to solve:
For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?
How do I solve it?
algebra-precalculus contest-math
New contributor
$endgroup$
I am practicing for a math contest and I encountered the following problem that I don't know how to solve:
For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?
How do I solve it?
algebra-precalculus contest-math
algebra-precalculus contest-math
New contributor
New contributor
edited 5 hours ago
NoChance
3,79221321
3,79221321
New contributor
asked 5 hours ago
swagbutton8swagbutton8
111
111
New contributor
New contributor
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
5 hours ago
1
$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
5 hours ago
$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
4 hours ago
1
$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
3 hours ago
1
$begingroup$
@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago
|
show 3 more comments
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
5 hours ago
1
$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
5 hours ago
$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
4 hours ago
1
$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
3 hours ago
1
$begingroup$
@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago
1
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
5 hours ago
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
5 hours ago
1
1
$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
5 hours ago
$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
5 hours ago
$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
4 hours ago
$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
4 hours ago
1
1
$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
3 hours ago
$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
3 hours ago
1
1
$begingroup$
@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago
$begingroup$
@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$
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add a comment |
$begingroup$
The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.
$endgroup$
add a comment |
$begingroup$
Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.
$endgroup$
add a comment |
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3 Answers
3
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3 Answers
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$begingroup$
Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$
$endgroup$
add a comment |
$begingroup$
Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$
$endgroup$
add a comment |
$begingroup$
Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$
$endgroup$
Hint:
$$x^2+5x+3=x+biff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $Delta=?$
answered 5 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
3,1671422
3,1671422
add a comment |
add a comment |
$begingroup$
The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.
$endgroup$
add a comment |
$begingroup$
The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.
$endgroup$
add a comment |
$begingroup$
The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.
$endgroup$
The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.
answered 4 hours ago
TojrahTojrah
6088
6088
add a comment |
add a comment |
$begingroup$
Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.
$endgroup$
add a comment |
$begingroup$
Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.
$endgroup$
add a comment |
$begingroup$
Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.
$endgroup$
Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.2k42061
42.2k42061
add a comment |
add a comment |
swagbutton8 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
5 hours ago
1
$begingroup$
Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant.
$endgroup$
– the_fox
5 hours ago
$begingroup$
A Parabola could have many tangent lines...However, it has only 1 min. value, may be this one is the one you are after. More here:desmos.com/calculator/zz7e1nwgf9 and here:info:math.stackexchange.com/questions/92165/…
$endgroup$
– NoChance
4 hours ago
1
$begingroup$
@NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though.
$endgroup$
– amd
3 hours ago
1
$begingroup$
@NoChance If you’re going to to that route, then you want the one that has the same slope as the given line.
$endgroup$
– amd
1 hour ago