Answer
$$
f(x)=2x^{2}-4x+5
$$
There are no $x$-intercepts.
The $y$-intercept is 5.
Vertex parabola is $\left(1 , 3\right)$.
The axis is $x=1$ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(1 , 3\right)$. gives the graph in Figure
Work Step by Step
$$
f(x)=2x^{2}-4x+5
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=2x^{2}-4x+5=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\begin{split}
\:x_{1,\:2} &=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\cdot \:5}}{2\cdot \:2} \\
&=\frac{4\pm \sqrt{-24}}{4}
\end{split}
$$
Since the radical is negative, then there are no $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(x)=2(0)^{2}-4(0)+5 =5.
$$
So the $y$-intercept is 5.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-4}{2.2}=\frac{4}{4}=1
$$
Substituting this into the equation gives
$$
f(x)=2(1)^{2}-4(1)+5=2-4+5=3
$$
The vertex is $\left(1 , 3\right)$.
The axis is $x=1 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(1 , 3\right)$. gives the graph in Figure