Answer
Vertex: $(-1,3)$
Opening: up
Shape: narrower
Vertical Axis
Discriminant: -24
Number of $x$-intercepts: 0
Work Step by Step
To find the vertex, the given function, $
f(x)=2x^2+4x+5
,$ 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)&=(2x^2+4x)+5
\\
f(x)&=2(x^2+2x)+5
.\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)&=2\left(x^2+2x+\left(\dfrac{2}{2}\right)^2\right)+\left[5-2\left(\dfrac{2}{2}\right)^2\right]
\\\\
f(x)&=2\left(x^2+2x+1\right)+\left[5-2\right]
\\
f(x)&=2\left(x+1\right)^2+3
.\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.
In the form $f(x)=a(x-h)^2+k$, the equation above is equivalent to
\begin{align*}
f(x)&=2\left(x-(-1)\right)^2+3
.\end{align*}
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 $
(-1,3)
$.
In the equation $
f(x)=2\left(x-(-1)\right)^2+3
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=2$ (greater than zero), then the parabola opens up.
With $a=2$ (greater than $1$), then the shape of the parabola is narrower than $y=x^2$.
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=
2
$, $b=
4
,$ and $c=
5
,$ then
\begin{align*}
b^2-4ac&\Rightarrow
4^2-4(2)(5)
\\&=
16-40
\\&=
-24
.\end{align*}
With a discriminant that is less than zero, then the parabola does not have $x$-intercepts.
