Find the x value of the stationary points of the graph y = x^2e^x

The problem here can be solved quite easily by thinking about the graph. A stationary point occurs on a graph when it becomes 'flat'. You can think of this visually as the top of a hill or the bottom of a ditch. Mathematically, this happens when the first derivative (dy/dx) is equal to 0 since at this point, the graph has a gradient of 0 (i.e. y will not increase or decrease as x changes). Using the product rule, we can find out dy/dx = 2xe^x + x^2e^x. Since we know that the gradient is 0, we can say that dy/dx = 2xe^x + x^2e^x = 0. To simplify this problem, we can put the expression into brackets such as e^x(2x+x^2) = 0. Therefore, either e^x or x^2 + 2x must equal 0. since e^x can never equal zero, this means that x^2 + 2x must equal 0. x^2 + 2x = x(x+2) = 0. Therefore, x = 0 or x = -2. The respective y values can be calculated by plugging these x values into y = x^2e^x.