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Help!
(x+2)(x+1)(2x-1) = 0
Show the point where the normal at x= -1 crosses the tangent at x=1 when x=0.68
I'm getting fairly close to this but can't seem to get it right.
Any help would be much appreciated.
Thanks
Show the point where the normal at x= -1 crosses the tangent at x=1 when x=0.68
I'm getting fairly close to this but can't seem to get it right.
Any help would be much appreciated.
Thanks
(x+2)(x+1)(2x-1)=0
2x^3 + 5x^2 + x - 2 = 0
y' = 6x^2 + 10x + 1
(x=-1)
y' = 6 - 10 + 1
y' = -3
therefore perp. = 1/3
y = 1/3x+c
and we know y = 0
0 = 1/3(-1) + c
c = 1/3
NORMAL: y = 1/3x + 1/3
y' = 6x^2 + 10x + 1
y' = 6 + 10 + 1 = 17
y = 17x + c
y = 2x^3 + 5x^2 + x - 2 = 6
6 = 17 + c
c = -11
TANGENT: y = 17x - 11
so:
17x - 11 = 1/3x + 1/3
16/2/2x = 11/1/3x
x = 0.68
----
hah, i even impressed myself
2x^3 + 5x^2 + x - 2 = 0
y' = 6x^2 + 10x + 1
(x=-1)
y' = 6 - 10 + 1
y' = -3
therefore perp. = 1/3
y = 1/3x+c
and we know y = 0
0 = 1/3(-1) + c
c = 1/3
NORMAL: y = 1/3x + 1/3
y' = 6x^2 + 10x + 1
y' = 6 + 10 + 1 = 17
y = 17x + c
y = 2x^3 + 5x^2 + x - 2 = 6
6 = 17 + c
c = -11
TANGENT: y = 17x - 11
so:
17x - 11 = 1/3x + 1/3
16/2/2x = 11/1/3x
x = 0.68
----
hah, i even impressed myself
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