AS- P1 differentiation (gradients) [closed]
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The curve with the equation y= ax^2 +bx +c passes through the point (1,2). The gradient of the curve is zero at the point (2,1). find the values of a, b and c.
derivatives
asked Jan 2 at 15:08
wizardwizard
204
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closed as off-topic by John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser Jan 3 at 4:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
The curve with the equation y= ax^2 +bx +c passes through the point (1,2). The gradient of the curve is zero at the point (2,1). find the values of a, b and c.
derivatives
asked Jan 2 at 15:08
wizardwizard
204
$endgroup$
closed as off-topic by John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser Jan 3 at 4:38
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." – John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The curve with the equation y= ax^2 +bx +c passes through the point (1,2). The gradient of the curve is zero at the point (2,1). find the values of a, b and c.
derivatives
asked Jan 2 at 15:08
wizardwizard
204
$endgroup$
The curve with the equation y= ax^2 +bx +c passes through the point (1,2). The gradient of the curve is zero at the point (2,1). find the values of a, b and c.
derivatives
derivatives
asked Jan 2 at 15:08
wizardwizard
204
asked Jan 2 at 15:08
wizardwizard
204
asked Jan 2 at 15:08
wizardwizard
204
asked Jan 2 at 15:08
wizardwizard
204
asked Jan 2 at 15:08
wizardwizard
204
204
closed as off-topic by John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser Jan 3 at 4:38
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." – John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser Jan 3 at 4:38
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." – John Doe, Keenan Kidwell, Lord Shark the Unknown, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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Hint: You are given that the points (1,2) and (2,1) are on the curve. This gives you two equations in terms of $a,b,c$ already. Then compute $frac{dy}{dx}$. You are given that this is zero when $x=2$, so with this, you get another equation in terms of $a,b,c$. Then solve these simultaneously. Can you do this?
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
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got it..thank you :)
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– wizard
Jan 2 at 15:59
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@wizard No worries!
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– John Doe
Jan 2 at 16:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: You are given that the points (1,2) and (2,1) are on the curve. This gives you two equations in terms of $a,b,c$ already. Then compute $frac{dy}{dx}$. You are given that this is zero when $x=2$, so with this, you get another equation in terms of $a,b,c$. Then solve these simultaneously. Can you do this?
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
$endgroup$
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
add a comment |
$begingroup$
Hint: You are given that the points (1,2) and (2,1) are on the curve. This gives you two equations in terms of $a,b,c$ already. Then compute $frac{dy}{dx}$. You are given that this is zero when $x=2$, so with this, you get another equation in terms of $a,b,c$. Then solve these simultaneously. Can you do this?
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
$endgroup$
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
add a comment |
$begingroup$
Hint: You are given that the points (1,2) and (2,1) are on the curve. This gives you two equations in terms of $a,b,c$ already. Then compute $frac{dy}{dx}$. You are given that this is zero when $x=2$, so with this, you get another equation in terms of $a,b,c$. Then solve these simultaneously. Can you do this?
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
$endgroup$
Hint: You are given that the points (1,2) and (2,1) are on the curve. This gives you two equations in terms of $a,b,c$ already. Then compute $frac{dy}{dx}$. You are given that this is zero when $x=2$, so with this, you get another equation in terms of $a,b,c$. Then solve these simultaneously. Can you do this?
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
answered Jan 2 at 15:23
John DoeJohn Doe
11.3k11239
11.3k11239
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
add a comment |
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
got it..thank you :)
$endgroup$
– wizard
Jan 2 at 15:59
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
$begingroup$
@wizard No worries!
$endgroup$
– John Doe
Jan 2 at 16:00
add a comment |
