Question

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=-4 x^{3}+6 x^{2}\)

Short answer

The graph of \(y=-4x^{3}+6x^{2}\) passes through the origin with local extrema at (0,0) and (1,2), and a point of inflection at (0.5, -0.5). As \(x\) approaches \(\pm\infty\), \(y\) also approaches \(-\infty\).

Step-by-step solution

Plotting Initial Points

Start with understanding the behaviour of the function as \(x\) approaches to infinity and negative infinity, and when \(x=0\). By substituting these values, it is easy to see that the graph will pass through point (0,0), and as \(x\) approaches to \(+\infty\) or \(-\infty\), \(y\) will also approach \(-\infty\).

Identifying Relative Extrema

Find the first derivative of the function, which is \(y' = -12x^{2} + 12x\). Set this equal to zero to find the \(x\) values at which a local maximum or minimum can occur. This gives \(x = 0\) and \(x = 1\). Substitute these values back into the original function to get the corresponding \(y\) values. Thus the relative extrema occur at points (0,0) and (1,2).

Identifying Inflection Points

Similarly, find the second derivative of the function, which would be \(y'' =-24x + 12\). Set this equal to zero for potential inflection point, which would be \(x = 0.5\). Substituting this value back into the original function, we get the coordinates of the point of inflection as (0.5, -0.5).

Sketching the Complete Graph

Now combine all these information to sketch the function. Remember that between the inflection point and relative extrema, the shape of the graph changes from concave down to concave up.