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2 Qs (P1)

Hm, I think I know the answer to this question but not sure and the second one I have no idea :redface:
Find the equation of the cruve thought (0,5) for why dy/dx = -2x + 3
The dy/dx is the gradient of the tangent. So when I use y-y1 = m(x-x1), do I take the gradient (m) to be -2?

2nd question: [After]
Use your diagram to solve the equation:
cosx = cos(x-60) 0<x<360
Indicate solutions clearly on graph.

How do I do this using a graph?
Thanks
rt6
Hm, I think I know the answer to this question but not sure and the second one I have no idea :redface:
Find the equation of the cruve thought (0,5) for why dy/dx = -2x + 3
The dy/dx is the gradient of the tangent. So when I use y-y1 = m(x-x1), do I take the gradient (m) to be -2?

dy/dx = -2x +3

int dy/dx:

y= -x^2 + 3x + c

we know that when x=0, y=5.

so 5 = 0 + 0 + c
c=5

=> y = -x^2 + 3x + 5
Reply 2
Avatar for lgs98jonee
lgs98jonee
2
rt6
Hm, I think I know the answer to this question but not sure and the second one I have no idea :redface:
Find the equation of the cruve thought (0,5) for why dy/dx = -2x + 3
The dy/dx is the gradient of the tangent. So when I use y-y1 = m(x-x1), do I take the gradient (m) to be -2?

2nd question: [After]
Use your diagram to solve the equation:
cosx = cos(x-60) 0<x<360
Indicate solutions clearly on graph.

How do I do this using a graph?
Thanks


2. Solutions will be where equations cross. This is at 30 and 210 i think.
Reply 3
Avatar for rt6
rt6
OP
1
Thanks for that. I thought I had to use the y-y1 = m(x-x1) equation :rolleyes:

Got it, thanks :smile:
Reply 4
Avatar for Newton
Newton
Sketch the graphs of cosx and cos(x-60) on the same graph, where cos(x-60) is the cosx graph but shifted 60deg to the right. Wherever they intersect those are the points where they are equal.

Newton.

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