finding a tangent line to a parabolaHow do I find the equation of a tangent line to a curve?Find intersection(s) between parametrized parabola and a lineIntersection point of tangent line with $X$ axisderive a parabola from two tangent linesWhy is this answer wrong? (point of intersection between parabola and line)Two questions on finding the equation of a parabola word problem- Klein's Calculus: An Intuitive and Physical ApproachEuclid 1999 Question 4(a) - Circle Tangent IntersectionFinding 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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finding a tangent line to a parabola
How do I find the equation of a tangent line to a curve?Find intersection(s) between parametrized parabola and a lineIntersection point of tangent line with $X$ axisderive a parabola from two tangent linesWhy is this answer wrong? (point of intersection between parabola and line)Two questions on finding the equation of a parabola word problem- Klein's Calculus: An Intuitive and Physical ApproachEuclid 1999 Question 4(a) - Circle Tangent IntersectionFinding 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
$begingroup$
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?
Edit:
I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.
algebra-precalculus contest-math
asked Apr 25 at 0:10
joenyjoeny
754
$endgroup$
|
show 4 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?
Edit:
I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.
algebra-precalculus contest-math
asked Apr 25 at 0:10
joenyjoeny
754
$endgroup$
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
7
$begingroup$
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
Apr 25 at 0:14
2
$begingroup$
@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
Apr 25 at 1:37
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
Apr 25 at 3:32
2
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55
|
show 4 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?
Edit:
I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.
algebra-precalculus contest-math
asked Apr 25 at 0:10
joenyjoeny
754
$endgroup$
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?
Edit:
I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.
algebra-precalculus contest-math
algebra-precalculus contest-math
asked Apr 25 at 0:10
joenyjoeny
754
asked Apr 25 at 0:10
joenyjoeny
754
edited Apr 25 at 18:03
joeny
asked Apr 25 at 0:10
joenyjoeny
754
asked Apr 25 at 0:10
joenyjoeny
754
asked Apr 25 at 0:10
joenyjoeny
754
754
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
7
$begingroup$
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
Apr 25 at 0:14
2
$begingroup$
@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
Apr 25 at 1:37
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
Apr 25 at 3:32
2
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55
|
show 4 more comments
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
7
$begingroup$
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
Apr 25 at 0:14
2
$begingroup$
@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
Apr 25 at 1:37
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
Apr 25 at 3:32
2
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55
1
1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
7
7
$begingroup$
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
Apr 25 at 0:14
$begingroup$
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
Apr 25 at 0:14
2
2
$begingroup$
@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
Apr 25 at 1:37
$begingroup$
@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
Apr 25 at 1:37
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
Apr 25 at 3:32
$begingroup$
@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
Apr 25 at 3:32
2
2
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55
|
show 4 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 $Delta=?$
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
$endgroup$
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.
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
$endgroup$
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.
answered Apr 25 at 0:21
TojrahTojrah
87010
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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 $Delta=?$
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
$endgroup$
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 $Delta=?$
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
$endgroup$
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 $Delta=?$
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
$endgroup$
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 Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
answered Apr 25 at 0:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
4,0431625
4,0431625
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.
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
$endgroup$
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.
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
$endgroup$
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.
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
$endgroup$
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 Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
answered Apr 25 at 0:21
Mohammad Riazi-KermaniMohammad Riazi-Kermani
42.4k42061
42.4k42061
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.
answered Apr 25 at 0:21
TojrahTojrah
87010
$endgroup$
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.
answered Apr 25 at 0:21
TojrahTojrah
87010
$endgroup$
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.
answered Apr 25 at 0:21
TojrahTojrah
87010
$endgroup$
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 Apr 25 at 0:21
TojrahTojrah
87010
answered Apr 25 at 0:21
TojrahTojrah
87010
answered Apr 25 at 0:21
TojrahTojrah
87010
answered Apr 25 at 0:21
TojrahTojrah
87010
87010
add a comment |
add a comment |
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1
$begingroup$
Do you know how to find a derivative?
$endgroup$
– R. Burton
Apr 25 at 0:13
7
$begingroup$
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
Apr 25 at 0:14
2
$begingroup$
@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
Apr 25 at 1:37
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
Apr 25 at 3:32
2
$begingroup$
If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.)
$endgroup$
– YawarRaza7349
Apr 25 at 3:55