Answer
Vertex: $\left(\dfrac{5}{2},\dfrac{37}{4}\right)$
Opening: down
Shape: same as $y=x^2$
Vertical Axis
Discriminant: 37
Number of $x$-intercepts: 2
Work Step by Step
To find the vertex, the given function, $
f(x)=-x^2+5x+3
,$ should be converted in the form $f(x)=a(x-h)^2+k$.
Grouping the variables together and making the coefficient of $x^2$ equal to $1$, the given function is equivalent to
\begin{align*}
f(x)&=(-x^2+5x)+3
\\
f(x)&=-(x^2-5x)+3
.\end{align*}
Completing the square of the expression with variables by adding $\left(\dfrac{b}{2}\right)^2,$ the expression above is equivalent to
\begin{align*}
f(x)&=-\left(x^2-5x+\left(\dfrac{-5}{2}\right)^2\right)+\left[3-3\left(\dfrac{-5}{2}\right)^2\right]
\\\\
f(x)&=-\left(x^2-5x+\dfrac{25}{4}\right)+\left[3-(-1)\left(\dfrac{25}{4}\right)\right]
\\\\
f(x)&=-\left(x-\dfrac{5}{2}\right)^2+\left[3+\dfrac{25}{4}\right]
\\\\
f(x)&=-\left(x-\dfrac{5}{2}\right)^2+\left[\dfrac{12}{4}+\dfrac{25}{4}\right]
\\\\
f(x)&=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}
.\end{align*}Note that $\left[a\left(\dfrac{b}{2}\right)^2\right]$ should be subtracted as well to negate the addition of that value when completing the square of the expression with variables.
Since the vertex of the quadratic function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
\left(\dfrac{5}{2},\dfrac{37}{4}\right)
$.
In the equation $
f(x)=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=-1$ (less than zero), then the parabola opens down.
With $|a|=|-1|=1$ (equal to $1$), then the shape of the parabola is the same as $y=x^2$.
Since $x$ is the squared variable in the equation, then the parabola has a vertical axis. Using $b^2-4ac$ to find the number of $x$-intercepts, with $a=
-1
$, $b=
5
,$ and $c=
3
,$ then
\begin{align*}
b^2-4ac&\Rightarrow
5^2-4(-1)(3)
\\&=
25+12
\\&=
37
.\end{align*}
With a discriminant that is greater than zero, then the parabola has two $x$-intercepts.
