91Ó°ÊÓ

The given function$f\left(x\right)=x^3+2x^2-5x-6$is of degree $3$.

The graph of f behaves like $y=x^3$ for large values of $\left|x\right|$.

For $x=0$, the function value is given as

$$\begin{gathered} f\left(0\right)=0^3+2·0^2-5·0-6 \\ f\left(0\right)=-6 \end{gathered}$$

So the y-intercept is $(0,-6)$.

Now set $f\left(x\right)=0$and solving for $x$we get

role="math" $\begin{array}{rcl}{x}^3+2x^2-5x-6& =& 0\\ x^3+x^2+x^2+x-6x-6& =& 0\\ x^2(x+1)+x(x+1)-6(x+1)& =& 0\\ (x+1)(x^2+x-6)& =& 0\\ (x+1)(x^2+3x-2x-6)& =& 0\\ (x+1)\left(x\right(x+3)-2(x+3\left)\right)& =& 0\\ (x+1)(x+3)(x-2)& =& 0\end{array}$

So the $x$intercepts are

$\begin{array}{rcl}x+1& =& 0\\ x& =& -1\end{array}$,

role="math" $\begin{array}{rcl}x+3& =& 0\\ x& =& -3\end{array}$

and

$\begin{array}{rcl}x-2& =& 0\\ x& =& 2\end{array}$

The zeros of the function are $-1,-3,2$. Each zero is of multiplicity $1$, so the graph of the function crosses the $x$axis at each zero.

The graph of the function can be given as

From the graph of f , we see that f has two turning points. Using MAXIMUM, one turning point is at $(-2.12,4.06)$, rounded to two decimal places. Using MINIMUM, the other turning point is at $(0.79,-8.21)$, rounded to two decimal places.

Plotting all the points and drawing a smooth curve we get

The graph is defined for every value of $x$, so its domain is set of all real numbers.

The range of the function is also the set of all real numbers.

Using the graph it can be seen that the graph is increasing in the interval $(-∞,-2.12)∪(0.79,∞)$.

The graph is decreasing in the intervalrole="math" localid="1646213516223" $(-2.12,0.79)$.