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finding a tangent line to a parabola
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finding a tangent line to a parabola
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraHow do I find the equation of a tangent line to a curve?Find intersection(s) between parametrized parabola and a lineHow to find equation of tangent line to $x^2 = 2y$ at $(-3, 9/2)$Find the equation of parabola tangent to a lineWhy is this answer wrong? (point of intersection between parabola and line)Euclid 1999 Question 4(a) - Circle Tangent Intersection4 tangent lines to parabola with points on circleFinding the Equation for the Line Tangent to a Parabola At a Given PointMirror image of the parabola about a tangentMaximizing $x^2 y + y^2 z + z^2 x − x^2 z − y^2 x − z^2 y$ for $x$, $y$, $z$ between $0$ and $1$ (inclusive)Find the numerical value of this expression
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 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
2 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
2 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
1 hour ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago
add a comment |
$begingroup$
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 2 hours ago
NoChance
3,79221321
3,79221321
New contributor
asked 2 hours ago
swagbutton8swagbutton8
61
61
New contributor
New contributor
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 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
2 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
2 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
1 hour ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago
add a comment |
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 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
2 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
2 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
1 hour ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago
1
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 hours ago
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
2 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
2 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
2 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
2 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
2 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
1 hour 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
1 hour ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago
add a comment |
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 $Delta=?$
$endgroup$
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 |
$begingroup$
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 |
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3 Answers
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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 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
3,1471422
3,1471422
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 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.2k42061
42.2k42061
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 2 hours ago
TojrahTojrah
5968
5968
add a comment |
add a comment |
swagbutton8 is a new contributor. Be nice, and check out our Code of Conduct.
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
2 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
2 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
2 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
1 hour ago
$begingroup$
@amd, thanks for your comment. I am not sure which tangent line one should pick then?
$endgroup$
– NoChance
29 mins ago