Answer
(a) $f(x) = (x-1)^2+2$
(b)
vertex: $(1, 2)$
x-intercepts: none
y-intercept: $3$
(c) Refer to the image below for the graph.
(d)
Domain $(-\infty, +\infty)$
Range: $[2, +\infty)$
Work Step by Step
$\bf{\text{ (a) Standard Form}}$
Group the terms with the variable $x$ to obtain:
$f(x) = (x^2-2x)+3$
Complete the square by adding $(-\frac{2}{2})^2=1$ inside the parentheses.
Subtract $(-\frac{2}{2})^2=1$ outside the parentheses.
$f(x) = (x^2-2x+1)+3-1$
Write the trinomial in factored form to obtain:
$\color{blue}{f(x) = (x-1)^2 + 2}$
$\bf{\text{(b) Vertex, x and y-intercepts}}$
The quadratic function has $a=1$ so the parabola opens upward.
The vertex is at $(1, 2)$, which is above the x-axis.
This means that the function has no x-intercepts as the graph will never touch the x-axis.
Solve for the y-intercept by solving for $f(0)$:
$f(x) = (x-1)^2+2
\\f(0) = (0-1)^2+2
\\f(0) = (-1)^2+2
\\f(0) = 1+2
\\f(0)=3$
Thus, the y-intercept is $3$.
$\bf{\text{(c) Graph}}$
The given function involves (i) a horizontal shift of 1 unit to the right and (ii) a vertical shift of 2 units upward of the parent function $y=x^2$.
Refer to the attached image in the answer part above for the graph.
$\bf{\text{(d) Domain and range}}$
The function is quadratic so its domain is the set of real numbers, $(-\infty, +\infty)$.
The y-values are from $2$ and above so the range is $(2, +\infty)$.
