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Cubic Curves problem

a) Can you find a cubic curve that has no stationary points.

That stationary points of a graph y= f(x) are those points (x,y) on the graph where f’(x)= 0. A stationary point on a cubic graph could either be stationary point of inflexion or a turning point.
The cubic equation f(x)=2x^3+3x^2+4x-5 has no stationary points.
Attached below is the graph for the cubic curve above and as you can see there are no stationary points apparent. But how can I prove this? I can differentiate this cubic into a quadratic equation in the form of ax^2+bx+c=0 which would give me 0=6x^2+6x+4 which I can either simplify by dividing both sides by two to give 0=3x^2+3x+2 and then use then substitute the values for a,b and c into the quadratic formula or alternatively use the quadratic formula at the outset. This would give an answer of x=(-3±i√15)/6 therefore meaning that there are no real roots or solutions for this equation thus proving that the cubic curve f(x)=2x^3+3x^2+4x-5 has no stationary points. Another way of looking at it is by solving the discriminant which is b^2-4ac=-15. A negative number, -15 in this case, has only one negative value for a cube root however it has no values for a square root therefore showing that there are no solutions for this equation.