C1 Solomon question Watch

PoliticallyDiverse
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#1
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Edit: Solved - post 5.
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Mark13
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You're on the right lines. You know the equation of the curve is y=x^2-3x+5, so you should be able to find dy/dx.

Now, dy/dx tells you the gradient of the tangent line to the curve at a given point. You need to find the gradient of the tangent line at the point Q - can you see how to do that?
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JRGC
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Differentiate equation of the curve, sub in the x value at Q. This will give you the gradient at Q also. So sub that and the coordinates of Q into equation of a straight line. Solve for C. Bingo, you have the equation.
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roar558
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(Original post by PoliticallyDiverse)
Hey everyone, I have the question (as below) and cannot figure out how you do question (c).

For a and b, respectively, I have the answers:

(a) P(1,3), Q(4,9)

(b) gradient = -1: y-3=-1(x-1)

y-3=-x+1

y=4-x

I've tried \frac{dy}{dx} for (c), but to no avail - could someone assist me, please?

for c), did you work out the gradient at using dy/dx at Q, and then plugged that value into the general equation of a straight line?
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PoliticallyDiverse
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(Original post by Mark13)
You're on the right lines. You know the equation of the curve is y=x^2-3x+5, so you should be able to find dy/dx.

Now, dy/dx tells you the gradient of the tangent line to the curve at a given point. You need to find the gradient of the tangent line at the point Q - can you see how to do that?
Wow, that seems like I've just missed something extremely simple...

Would it be:

y=x^2 - 3x -5

\frac{dy}{dx} = 2x - 3

2(4)-3 = 5

y-9=5(x-4)

y-4=5x-20

y=5x-11?
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MostCompetitive
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(Original post by PoliticallyDiverse)
Hey everyone, I have the question (as below) and cannot figure out how you do question (c).

For a and b, respectively, I have the answers:

(a) P(1,3), Q(4,9)

(b) gradient = -1: y-3=-1(x-1)

y-3=-x+1

y=4-x

I've tried \frac{dy}{dx} for (c), but to no avail - could someone assist me, please?

Method:

1) differentiate the equation of the curve
2) substitute the x co-ordinate of the point you want to find the tangent at ( x co-ordinate of Q in this case) into \frac{dy}{dx}
3) What you will get is the gradient of the tangent. Use y-y_1=m(x-x_1), where x_1 is the x co-ordinate, y_1 is the y co-ordinate and m is the gradient to find the equation of the tangent.
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MostCompetitive
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(Original post by PoliticallyDiverse)
Wow, that seems like I've just missed something extremely simple...

Would it be:

y=x^2 - 3x -5

\frac{dy}{dx} = 2x - 3

2(4)-3 = 5

y-9=5(x-4)

y-4=5x-20

y=5x-11?
Well done.
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Nomolos
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#8
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What exactly has it got to do with Solomon?
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PoliticallyDiverse
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#9
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(Original post by joshgoldman)
What exactly has it got to do with Solomon?
It's from a Solomon paper...
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Nomolos
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#10
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(Original post by PoliticallyDiverse)
It's from a Solomon paper...
There's a board called Solomon? Coooooool.
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