- The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where
$$f ( x ) = a x ^ { 3 } + 15 x ^ { 2 } - 39 x + b$$
and \(a\) and \(b\) are constants.
Given
- the point \(( 2,10 )\) lies on \(C\)
- the gradient of the curve at \(( 2,10 )\) is - 3
- (i) show that the value of \(a\) is - 2
(ii) find the value of \(b\). - Hence show that \(C\) has no stationary points.
- Write \(\mathrm { f } ( x )\) in the form \(( x - 4 ) \mathrm { Q } ( x )\) where \(\mathrm { Q } ( x )\) is a quadratic expression to be found.
- Hence deduce the coordinates of the points of intersection of the curve with equation
$$y = \mathrm { f } ( 0.2 x )$$
and the coordinate axes.