College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 36

Identify the type of curve which is represented by the equation $3x^2 - (6x + y) = 10 $
Find the foci and vertices(if any), and sketch the graph

$
\begin{equation}
\begin{aligned}
3x^2 - (6x + y) &= 10 && \text{Distribute } 6\\
\\
3x^2 - 6x - 6y &= 10 && \text{Add } 6y\\
\\
3x^2 - 6x &= 10 + 6y && \text{Factor 3 from the right}\\
\\
3(x^2 - 2x) &= 10 + 6y && \text{Divide by 3 both sides}\\
\\
x^2 - 2x &= \frac{10}{3} + 2y && \text{Complete the square; Add } \left( \frac{-2}{2} \right)^2 = 1\\
\\
x^2 - 2x + 1 &= \frac{10}{3} + 2y + 1 && \text{Perfect Square}\\
\\
(x - 1)^2 &= 2y + \frac{13}{3} && \text{Factor out 2 from the right}\\
\\
(x -1)^2 &= 2 \left( y + \frac{13}{6} \right)
\end{aligned}
\end{equation}
$

The equation is parabola that has form $(x-h)^2 = 4p(y-k)$ with center at $(h,k)$ and opens upward. The graph is obtained from $x^2 = 4py$ by shifting 1 unit
to the right and $\displaystyle \frac{13}{6}$ units downward. And since $4p = 2$, we have $\displaystyle p = \frac{1}{2}$. It means that the focus is
$\displaystyle \frac{1}{2}$ units above the vertex. Therefore, the following is determined as

$
\begin{equation}
\begin{aligned}
\text{vertex } & \left( 1, -\frac{13}{6} \right)\\
\\
\text{foci } & \left( 1, -\frac{13}{6} + \frac{1}{2} \right) && \rightarrow && \left( 1, \frac{-5}{3} \right)
\end{aligned}
\end{equation}
$

Then, the graph is