Chapter 4, Problem 15RE

To determine

To analyze:the givenpolynomial function

Answer to Problem 15RE

  

Explanation of Solution

Given:

  $f\left(x\right)=-2x^3+4x^2$

Calculation:

The given function is $f\left(x\right)=-2x^3+4x^2$

Step 1: End Behaviour: the graphfresembles that of the power function $y=x^3$ for largervalues of $\left|x\right|$ .

Step 2: The y intercept is. $f\left(0\right)=-2{\left(0\right)}^3+4{\left(0\right)}^2=0$ The x intercepts satisfy the equation

  $f\left(x\right)=-2x^3+4x^2=-2x^2\left(x-2\right)=0$

So

  $-2x^2=0$ Or $x-2=0$

  $x=0$ Or $x=2$

The x -intercepts are 0 and 2.

Step 3: The intercept 0 is zero of multiplicity 2, so the graph of fwill touch x -axis at 0; 2 is zero with multiplicity 1, so graph of f will cross x -axis at 2.

Step 4: The graph of f will contain at most two turning points.

Step 5: The two x -intercepts are -4 and 2.

Near 0: $f\left(x\right)=-2x^3+4x^2=-2x^2\left(x-2\right)\approx -2x^2\left(0-2\right)=4x^2$ A parabola opening up

Near 2: $f\left(x\right)=-2x^3+4x^2=-2x^2\left(x-2\right)\approx -2{\left(2\right)}^2\left(x-2\right)=-8\left(x-2\right)$ A line with slope -8

Step 6: After combining steps 1 through 5 we get graph as shown below:

  

Conclusion:

Therefore, the polynomial function $f\left(x\right)=-2x^3+4x^2$ is analysed with graph.