Answer
Choice C
Work Step by Step
To find the vertex of the given equation, $
y=-\dfrac{1}{2}x^2-x+1
,$ convert the equation in the form $y=a(x-h)^2+k$.
Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
y&=\left(-\dfrac{1}{2}x^2-x\right)+1
\\
y&=-\dfrac{1}{2}\left(x^2+2x\right)+1
.\end{align*}
Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to
\begin{align*}
y&=-\dfrac{1}{2}\left(x^2+2x+\left(\dfrac{2}{2}\right)^2\right)+\left[1-\left(-\dfrac{1}{2}\right)\left(\dfrac{2}{2}\right)^2\right]
\\\\
y&=-\dfrac{1}{2}\left(x^2+2x+1\right)+\left[1+\dfrac{1}{2}\right]
\\\\
y&=-\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
-\dfrac{1}{2}\left(\dfrac{2}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of the equation $y=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic equation above is $
\left(-1,\dfrac{3}{2}\right)
$. Thus, the vertex is located on the second quadrant.
In the equation $
y=-\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=-\dfrac{1}{2}$ (less than zero), then the parabola opens downwards.
With the vertex located at Quadrant $2$ and the parabola opens down, then the corresponding graph of the given equation is Choice C.
