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C1 Solomon question

Edit: Solved - post 5.
(edited 12 years ago)
Reply 1
Avatar for Mark13
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?
Reply 2
Avatar for JRGC
JRGC
7
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.
Reply 3
Avatar for roar558
roar558
11
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)P(1,3), Q(4,9)

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

y−3=−x+1y-3=-x+1

y=4−xy=4-x

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



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?
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... :colondollar:

Would it be:

y=x2−3x−5y=x^2 - 3x -5

dydx=2x−3\frac{dy}{dx} = 2x - 3

2(4)−3=52(4)-3 = 5

y−9=5(x−4)y-9=5(x-4)

y−4=5x−20y-4=5x-20

y=5x−11y=5x-11? :smile:
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)P(1,3), Q(4,9)

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

y−3=−x+1y-3=-x+1

y=4−xy=4-x

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



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 dydx\frac{dy}{dx}
3) What you will get is the gradient of the tangent. Use y−y1=m(x−x1)y-y_1=m(x-x_1), where x1x_1 is the x co-ordinate, y1y_1 is the y co-ordinate and m is the gradient to find the equation of the tangent.
Original post by PoliticallyDiverse
Wow, that seems like I've just missed something extremely simple... :colondollar:

Would it be:

y=x2−3x−5y=x^2 - 3x -5

dydx=2x−3\frac{dy}{dx} = 2x - 3

2(4)−3=52(4)-3 = 5

y−9=5(x−4)y-9=5(x-4)

y−4=5x−20y-4=5x-20

y=5x−11y=5x-11? :smile:


Well done.
Reply 7
Avatar for Nomolos
Nomolos
12
What exactly has it got to do with Solomon?
Original post by joshgoldman
What exactly has it got to do with Solomon?


It's from a Solomon paper...
Reply 9
Avatar for Nomolos
Nomolos
12
Original post by PoliticallyDiverse
It's from a Solomon paper...


There's a board called Solomon? Coooooool.

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