154480

A curve with no apparent stationary points is completely possible and occurs when no point on the curve has a gradient of zero. Having plotted some cubic curves prior to this, I decided to find any quadratic which doesn't factorise easily, for example you'd have to either complete the square or its non-factorable. I came up with a few quadratic expressions: this includes 6x^2+24x. There are many more expressions I could have come up with however the expression listed above seemed the most likely to work. It made sense for me to test. From here I needed to differentiate the quadratic expression in order to produce a cubic expression that I am able to plot by hand and check to see whether or not there are any stationary points on the curve. I integrated the expression 6x^2+24x in order to give 2x^3+24x and this plotted a curve with no stationary points (Attached below is a word document with the image of that expression plotted). This is because dy/dx= 6x^2+24x. If you substitute a value of zero for the gradient this gives 6x^2+24=0. 6x^2=-24. x^2=-4 which there are no solutions for since a negative number has no real square roots and only one -ve cube root. For this reason, the expression 6x^2+24 when integrated gives a cubic expression when plotted produces no stationary points.