Answer
$$
y=1.1 -0.3x-0.3 x^{2}
$$
(a)
The critical number is : $\frac{-1}{2}$,
(b)
The function is increasing on interval $(-\infty, \frac{-1}{2} ),$
(c)
The function is decreasing on interval $( \frac{-1}{2} ,\infty ).$
Work Step by Step
$$
y=1.1 -0.3x-0.3 x^{2}
$$
First, find the points where the derivative $y^{\prime }$ is $0$
Here
$$
\begin{aligned}
y^{\prime}(x) &=-0.3-0.3(2) x\\
&=-0.3-0.6 x\\
\end{aligned}
$$
Solve the equation $y^{\prime}(x) =0 $ to get
$$
\begin{aligned}
y^{\prime}(x) & =-0.3-0.6 x=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
-0.6 x &=0.3 \\
x &=\frac{0.3}{-0.6 } \\
&=\frac{-1}{2}\\
\end{aligned}
$$
(a)
The critical number is : $\frac{-1}{2}$.
Now, we can use the first derivative test.
Check the sign of $y^{\prime}(x)$ in the intervals
$$
(-\infty, \frac{-1}{2} ), \quad ( \frac{-1}{2},\infty).
$$
(1)
Test a number in the interval $(-\infty, \frac{-1}{2} )$ say $-1$:
$$
\begin{aligned}
y^{\prime}(-1) &=-0.3-0.6 (-1) \\
&=0.3 \\
&\gt 0
\end{aligned}
$$
to see that $ y^{\prime}(x)$ is positive in that interval, so $y(x)$ is increasing on $(-\infty, \frac{-1}{2} )$
(2)
Test a number in the interval $( \frac{-1}{2} , \infty )$ say $0$:
$$
\begin{aligned}
y^{\prime}(0)& =-0.3-0.6 (0) \\
&=-0.3 \\
& \lt 0
\end{aligned}
$$
to see that $ y^{\prime}(x)$ is negative in that interval, so $y(x)$ is decreasing on $( \frac{-1}{2} , \infty ).$
So,
(b) The function is increasing on interval $(-\infty, \frac{-1}{2} )$
(c) The function is decreasing on interval $( \frac{-1}{2} ,\infty )$
