91Ó°ÊÓ

The function is,

$f\left(x\right)=3x^2-2x-5$.

The first, second, and third-order Maclaurin polynomials, that is, $P_1\left(x\right),P_2\left(x\right),P_3\left(x\right)$are,

$$\begin{gathered} P_1\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x \\ {P}_2\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2 \\ {P}_3\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3 \end{gathered}$$

Finding the value at $x=0$,

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

Finding the derivatives of the function,

role="math" localid="1649585912508" $$\begin{gathered} f\text{'}\left(x\right)=\frac{d(3x^2-2x+5)}{dx} \\ =3\frac{d\left(x^2\right)}{dx}-2\frac{dx}{dx}+5\frac{d0}{dx} \\ =6x-2 \\ f\text{'}\left(0\right)=6\left(0\right)-2 \\ =-2 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(6x-2)}{dx} \\ =6\frac{dx}{dx} \\ =6 \\ f\text{'}\text{'}\left(0\right)=6 \end{gathered}$$

Also,

role="math" localid="1649586030408" $$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d6}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\left(0\right)=0 \end{gathered}$$

Therefore, the first, second and third-order Maclaurin polynomials are,

$$\begin{gathered} P_1\left(x\right)=5-2x \\ {P}_2\left(x\right)=5-2x+3x^2 \\ {P}_3=5-2x+3x^2 \end{gathered}$$

Here,$P_2\left(x\right)=P_3\left(x\right)$. This is because for any polynomial function $f$of degree $n$, the $nth$Maclaurin Polynomial is, $P_m\left(x\right)=f\left(x\right)wheremâ‰\yen n$.

The graph for $f\left(x\right),P_1\left(x\right),P_2\left(x\right)$is,