Early differentiation question help please

[From: ] [author: ] [Date: 11-11-11] [Hit: ]

-2/2x = 2 , -1y = x + 2y = 4 , 1(2 , 4) and (-1 , 1)Now, find the tangent lines to y = x^2 at those points:y = x^2dy/dx = 2xx = 2dy/dx = 2 * 2 = 4y - 4 = 4 * (x - 2)y = 4 * (x - 2) + 4y = 4x - 8 + 4y = 4x - 4y = 4 * (x - 1)x = -1dy/dx = -2y - 1 = -2 * (x - (-1))y - 1 = -2 * (x + 1)y - 1 = -2x - 2y = -2x - 1Now find when the 2 lines intersect:4 * (x - 1) = -2x - 14x - 4 = -2x - 16x = 3x = 3/6x = 1/2y = -2x - 1y = -1 - 1y = -2(1/2 ,......

you have to find the point at which the tangent of y = x ^ 2 ( which is 2x) meets the line y = x + 2, so,
set 2x equal to x + 2 to get the x value of the point,
2x=x+2
x=2
Insert 2 into y = x + 2 to get y
y=4
point: (2,4)

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First, find when y = x^2 intersects with y = x + 2

x^2 = x + 2
x^2 - x - 2 = 0
x = (1 +/- sqrt(1 + 8)) / 2
x = (1 +/- 3) / 2
x = 4/2 , -2/2
x = 2 , -1

y = x + 2
y = 4 , 1

(2 , 4) and (-1 , 1)

Now, find the tangent lines to y = x^2 at those points:

y = x^2
dy/dx = 2x

x = 2
dy/dx = 2 * 2 = 4

y - 4 = 4 * (x - 2)
y = 4 * (x - 2) + 4
y = 4x - 8 + 4
y = 4x - 4
y = 4 * (x - 1)

x = -1
dy/dx = -2

y - 1 = -2 * (x - (-1))
y - 1 = -2 * (x + 1)
y - 1 = -2x - 2
y = -2x - 1

Now find when the 2 lines intersect:

4 * (x - 1) = -2x - 1
4x - 4 = -2x - 1
6x = 3
x = 3/6
x = 1/2

y = -2x - 1
y = -1 - 1
y = -2

(1/2 , -2) is the point of intersection

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