Answer
(a)
The critical number is : $\frac{17}{12}$,
(b) The function is increasing in interval $(-\infty, \frac{17}{12} ),$
(c) The function is decreasing in interval $(\frac{17}{12} ,\infty ).$
Work Step by Step
$$
y=2.3 +3.4x-1.2 x^{2}
$$
First, find the points where the derivative $y^{\prime }$ is $0$
Here
$$
\begin{aligned}
y^{\prime}(x) &=3.4-1.2 (2)x\\
&=3.4-2.4 x\\
\end{aligned}
$$
Solve the equation $y^{\prime}(x) =0 $ to get
$$
\begin{aligned}
y^{\prime}(x) & =3.4-2.4 x=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
3.4-2.4 x=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
-2.4 x &=-3.4 \\
x &=\frac{-3.4}{-2.4}=\frac{17}{12}\\
\end{aligned}
$$
(a)
The critical number is : $\frac{17}{12}$.
Now, we can use the first derivative test.
Check the sign of $y^{\prime}(x)$ in the intervals
$$
(-\infty, \frac{17}{12} ), \quad (\frac{17}{12},\infty).
$$
(1)
Test a number in the interval $(-\infty, \frac{17}{12} )$ say $0$:
$$
\begin{aligned}
y^{\prime}(0) =3.4-2.4 (0) \\
&=3.4 \\
&\gt 0
\end{aligned}
$$
to see that $ y^{\prime}(x)$ is positive in that interval, so $y(x)$ is increasing on $(-\infty, \frac{17}{12} )$
(2)
Test a number in the interval $(\frac{17}{12} , \infty )$ say $2$:
$$
\begin{aligned}
y^{\prime}(2) =3.4-2.4 (2) \\
&=3.4-4.8 \\
&=-1.4 \\
& \lt 0
\end{aligned}
$$
to see that $ y^{\prime}(x)$ is negative in that interval, so $y(x)$ is decreasing on $(\frac{17}{12} , \infty ).$
(b) The function is increasing in interval $(-\infty, \frac{17}{12} )$
(c) The function is decreasing in interval $(\frac{17}{12} ,\infty )$
