the curve intersects the x-axis at (2,0)
y'(x) = -20/(2x+1)^2
y'(2) = -4/5
tangent at (2,0) is
y = -4/5 (x-2)
intersects the y-axis at (0,8/5)
AC is thus
y = (x-2)(-4/5)
5y = 8-4x
5y+4x=8
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y = 1/6 (2x-3)^3 - 4x
y' = (2x-3)^2 - 4
the curve has y-intercept at (0,-9/2)
y'(0) = 5
so, the line is
y+9/2 = 5x
1. The curve y=10/2x+1 -2 intersects the x-axis at A. The tangent to the curve at A intersects the y-axis at C.
(i) Show that the equation of AC is 5y+4x=8
(ii) Find the distance of AC
2. The equation of a curve is y=1/6(2x-3)^3-4x
(i) Find dy/dx
(ii) Find the equation of the tangent to the curve at the point where the curve intersects the y-axis.
1 answer
1 answer