How do you sketch the curve y=(x^2 - 4)(x+3), marking on turning points and values at which it crosses the x axis
First, factorise the equation into y = (x+3)(x-2)(x+2), nothing that x2-4 is the difference of two squares so is easy to factorise. From the factorised equation, the points at which the curve crosses the x axis can be identified, by setting the equation equal to 0. The x values that make each bracket 0 will give the x values where the curve crosses the x axis, which in this case is (-3, 0) (-2, 0) and (2,0). Mark these points on an x-y axis.
Secondly, multiply out the brackets of the equation to give y=x3 + 3x2-4x-12. The highest power is x3 and it is positive, which means it will take the shape of a common x3 graph, starting in the lower left quadrant for very large negative x values and ending in the upper right quadrant for very large positive x values . Differentiate the equation to find the x values of the turning points of the curve, and substitute these values back into the original equation to get the corresponding y values. (Differentiation with respect to x gives 0=3x2+6x-4, solve using the quadratic formula to give turning points). Plot turning points on axis, then use all the information to sketch the curve.
