How could I sketch a graph of y=2x^3-3x^2?

To sketch a graph of an equation, there are several key features to consider. Firstly, we can consider where the graph crosses the x- and y-axes. When x=0, y=0, so the graph goes through the origin. If y=0, x=0 is possible, but as y=x^2(2x-3), we could also have x=3/2.
This tells us where the graph crosses the axes, so now we can ask where its stationary points are. We can find that the derivative of y is 6x^2-6x, and this is equal to zero when x=0 or x=1. To find out how the function behaves in general, think about what would happen if x is very large: x^3 will get much bigger than x^2, as it has an extra factor of x, and so for very large x, y is large and positive. This gives us enough information to find the shape of the graph.
We have found two turning points, and a cubic cannot have more than two turning points, and so we can tell from this which direction the graph will go between the turning points, and then sketch the graph.

LH

Related Maths A Level answers

All answers ▸

Find the x-values of the turning points on the graph, y=(3-x)(x^2-2)


b) The tangent to C at P meets the coordinate axes at the points Q and R. Show that the area of the triangle OQR, where O is the origin, is 9/(3-e)


A curve has the equation 2x^2 + xy - y^2 +18 = 0. (1) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis.


A curve C has equation: x^3+2xy-x-y^3-20=0. Find dy/dx in terms of x and y.