Chapter 4.5, Problem 2E

To determine

To find: the different solution from the given questions.

Answer to Problem 2E

  $y$ intercept of the graph is $\left(0,2\right)$ .

  $x$ intercept of the graph is $\left(-1,0\right),\left(1,0\right)$ and $\left(2,0\right)$ .

The real solution are $x=-1,x=1$ and $x=2$ .

The zeros are $x=-1,x=1$ and $x=2$ .

Explanation of Solution

Given information:

The y -intercepts of the graph of $y=x^3-2x^2-x+2$ .

The x -intercepts of the graph of $y=x^3-2x^2-x+2$ .

The real solutions of $x^3-2x^2-x+2=0$ .

The real zeros of $f\left(x\right)=x^3-2x^2-x+2$ .

Calculation:

To find the $y$ intercept of $y=x^3-2x^2-x+2$ for $x=0$ .

  $\begin{array}{c}y={\left(0\right)}^3-2{\left(0\right)}^2+2\\ =2\end{array}$

Therefore, $y$ intercept of the graph is $\left(0,2\right)$ .

To find the $x$ intercept of $y=x^3-2x^2-x+2$ , set $y=0$ .

  $\begin{array}{c}{x}^3-2x^2-x+2=0\\ \left(x^3-2x^2\right)+\left(-x+2\right)=0\\ x^2\left(x-2\right)-1\left(x-2\right)=0\\ \left(x^2-1\right)\left(x-2\right)=0\\ \left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ x=-1,1,2\end{array}$

Therefore, $x$ intercept of the graph is $\left(-1,0\right),\left(1,0\right)$ and $\left(2,0\right)$ .

To find the all real solutions of $y=x^3-2x^2-x+2$ , set $y=0$ ,

  $\begin{array}{c}{x}^3-2x^2-x+2=0\\ \left(x^3-2x^2\right)+\left(-x+2\right)=0\\ x^2\left(x-2\right)-1\left(x-2\right)=0\\ \left(x^2-1\right)\left(x-2\right)=0\\ \left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ x=-1,1,2\end{array}$

Therefore, the real solution are $x=-1,x=1$ and $x=2$ .

To find the real zeros of $y=x^3-2x^2-x+2$ , set $y=0$ .

  $\begin{array}{c}{x}^3-2x^2-x+2=0\\ \left(x^3-2x^2\right)+\left(-x+2\right)=0\\ x^2\left(x-2\right)-1\left(x-2\right)=0\\ \left(x^2-1\right)\left(x-2\right)=0\\ \left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ x=-1,1,2\end{array}$

Therefore, the zeros are $x=-1,x=1$ and $x=2$ .

Thus, finding $y$ intercepts is different and finding real solution, zeros and $x$ intercepts are same.