Answer
Critical points $x=-1$ and $x\approx3.16$, no endpoints.
At $x=-1$, $y=4$ local maximum.
At $x\approx3.16$, $y=\approx-3.1$ local minimum.
Work Step by Step
Step 1. Given the function
$y=\begin{cases} -x^2/4-x/2+15/4\hspace1cm x\leq1 \\ x^3-6x^2+8x\hspace2.2cm x\gt1 \end{cases}$,
we can find its derivative as $y'=\begin{cases} -x/2-1/2\hspace1.4cm x\leq1 \\ 3x^2-12x+8\hspace1cm x\gt1 \end{cases}$.
We have $ \lim_{x\to1^+}f'(x)=-1, \lim_{x\to1^-}f'(x)=-1$ and $f'(1)=-1$ is defined.
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=-1$ and the solution $3x^2-12x+8=0, x\gt1$ which gives $x=\frac{2+2\sqrt 3}{3}\approx3.16$. The domain of the function is $(-\infty,\infty)$ and there are no endpoints.
Step 3. At $x=-1$, $y= -(-1)^2/4-(-1)/2+15/4=4$ and we can identify this as a local maximum (use local test points as necessary) as shown in the figure.
Step 4. At $x=\frac{2+2\sqrt 3}{3}\approx3.16$, $y= (\frac{2+2\sqrt 3}{3})^3-6(\frac{2+2\sqrt 3}{3})^2+8(\frac{2+2\sqrt 3}{3})\approx-3.1$ and we can identify this as a local minimum (use local test points as necessary) as shown in the figure.