Answer
Vertex: $\left(\dfrac{7}{2},\dfrac{57}{4}\right)$
Opening: down
Shape: same as $y=x^2$
Vertical Axis
Discriminant: 57
Number of $x$-intercepts: 2
Work Step by Step
To find the vertex, the given function, $
f(x)=-x^2+7x+2
,$ 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+7x)+2
\\
f(x)&=-(x^2-7x)+2
.\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^2-7x+\left(\dfrac{-7}{2}\right)^2\right)+\left[2-(-1)\left(\dfrac{-7}{2}\right)^2\right]
\\\\
f(x)&=-\left(x^2-7x+\dfrac{49}{4}\right)+\left[2+\dfrac{49}{4}\right]
\\\\
f(x)&=-\left(x-\dfrac{7}{2}\right)^2+\left[\dfrac{8}{4}+\dfrac{49}{4}\right]
\\\\
f(x)&=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{57}{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{7}{2},\dfrac{57}{4}\right)
$.
In the equation $
f(x)=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{57}{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
$, $b=
7
,$ and $c=
2
,$ then
\begin{align*}
b^2-4ac&\Rightarrow
7^2-4(-1)(2)
\\&=
49+8
\\&=
57
.\end{align*}
With a discriminant that is greater than zero, then the parabola has two $x$-intercepts.
