Question

graph f(x)=-x^{3}-x^{2}+2x - 3 and give points

Explanation:

Step1: Find the y - intercept

Set \(x = 0\) in \(y=-x^{3}-x^{2}+2x - 3\).
\(y=-0^{3}-0^{2}+2\times0 - 3=-3\). So the point is \((0,-3)\).

Step2: Find some other points

Let \(x = 1\), then \(y=-1^{3}-1^{2}+2\times1 - 3=-1 - 1+2 - 3=-3\). So the point is \((1,-3)\).
Let \(x=-1\), then \(y=-(-1)^{3}-(-1)^{2}+2\times(-1)-3 = 1 - 1-2 - 3=-5\). So the point is \((-1,-5)\).
Let \(x = 2\), then \(y=-2^{3}-2^{2}+2\times2 - 3=-8 - 4 + 4-3=-11\). So the point is \((2,-11)\).
Let \(x=-2\), then \(y=-(-2)^{3}-(-2)^{2}+2\times(-2)-3=8 - 4-4 - 3=-3\). So the point is \((-2,-3)\).

To graph the function \(y =-x^{3}-x^{2}+2x - 3\), we can plot the points \((0,-3)\), \((1,-3)\), \((-1,-5)\), \((2,-11)\), \((-2,-3)\) and then sketch the curve. The general shape of a cubic function \(y = ax^{3}+bx^{2}+cx + d\) (here \(a=-1\lt0\)): as \(x\to+\infty\), \(y\to-\infty\) and as \(x\to-\infty\), \(y\to+\infty\).

Answer:

Points: \((0,-3)\), \((1,-3)\), \((-1,-5)\), \((2,-11)\), \((-2,-3)\)