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Given that $f(x) = 2x^{3} - 5x^{2} - 4x + 3$ and the intercepts of $f(x)$ are $(-1;0), (\frac{1}{2};0)$ and $(3;0)$

  1. Determine the coordinates of the turning points (stationary points) and the point of inflection of $f(x)$
  2. Draw a neat graph of $f(x)$. Clearly indicate the coordinates of the intercepts with the axes as well as the coordinates of the turning and inflection points.
  3. Detyermine the equation of the tangent to the graph of $f(x) = 2x^{3} - 5x^{2} - 4x +3 $ at the point where $x=1$
in Grade 12 Maths by Diamond (37.9k points) | 13 views

1 Answer

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1. dy/dx = 6x^2 -10x -4 = 0 at stationary points

               (6x+2)(x-2)=0

                         x = -1/3 or 2

substitute x values into cubic function, y = 3.7 or -9

= (-1/3;3.7) and (2;-9)

Point of inflection = (-1/3 + 2)/2 ; (3.7 +(-9))/2 = (0.83;-2.65)

2. Let x=0, y=3 for y-intercept

    x-intercepts are given

 Coordinates of turning points and inflection already calculated from 1.

 The cubic graph will decrease from left, increase and decrease, as coefficient of x^3 is positive.

3. dy/dx at x = 1, = 6(1)^2 - 10(1)-4

                            = 6-10-4

                           = -8

substituting x=1 into cubic graph, y= -4

hence y-(-4)= -8(x-1)

                y= -8x +4
by Silver Status (28.5k points)

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