Answer
Choice A
Work Step by Step
To find the vertex of the given equation, $
y=-x^2+3x+5
,$ 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&=(-x^2+3x)+5
\\
y&=-(x^2-3x)+5
.\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&=-\left(x^2-3x+\left(\dfrac{3}{2}\right)^2\right)+\left[5-(-1)\left(\dfrac{3}{2}\right)^2\right]
\\\\
y&=-\left(x^2-3x+\dfrac{9}{4}\right)+\left[5+\dfrac{9}{4}\right]
\\\\
y&=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{29}{4}
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
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(\dfrac{3}{2},\dfrac{29}{4}\right)
$. Thus, the vertex is located on the first quadrant.
In the equation $
y=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{29}{4}
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=-1$ (less than zero), then the parabola opens downwards.
With the vertex located at Quadrant $1$ and the parabola opens down, then the corresponding graph of the given equation is Choice A.
