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

  1. Determine the co-ordinates of the turning points of $f$.
  2. Draw a neat sketch graph of $f$. Clearly indicate the co-ordinates of the intercepts with the axes, as well as the co-ordinates of the turning points.
  3. For which values of $k$ will the equation $f(x) = k$ , have exactly two real roots?
  4. Determine 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 Bronze Status (5.9k points) | 18 views

1 Answer

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

                (3x+1)(x-2)=0

               x= -1/3  or 2

substituting into cubic function, y=3.70 or -9, (-1/3 ; 3.70) and (2 ; -9) as turning points

2. y-intercept, x=0, y=3, (0;3)

Then plot with x-intercepts given and turning points, graph decreases from left, increase and lastly decreases

4. substitute x=1 into first derivative

          = 6-10-4

            = -8

x=1, y= -4 , x substituted from original function to give (1; -4)

y-y= m(x-x)

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

   y= -8x + 4
by Diamond (82.9k points)
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