Monday

March 2, 2015

March 2, 2015

Posted by **Stephen** on Monday, January 14, 2013 at 11:33pm.

(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.

- Math -
**Steve**, Tuesday, January 15, 2013 at 12:00pmthe 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

-------------------------------

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

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