Q. Which is the graph of \(f(x) = x^2 – 2x + 3\)?

Problem: Analyze the quadratic function \( f(x)=x^{2}-2x+3 \)

Step 1 – Identify the coefficients

The function is in standard form \( f(x)=ax^{2}+bx+c \). Here
\[
a=1,\qquad b=-2,\qquad c=3.
\]

Step 2 – Determine the direction of the parabola

Since the leading coefficient \( a=1 \) is positive, the parabola opens upward.

Step 3 – Find the vertex

The \(x\)-coordinate of the vertex is given by
\[
x=-\dfrac{b}{2a}=-\dfrac{-2}{2(1)}=\dfrac{2}{2}=1.
\]
Evaluate \(f\) at \(x=1\) to get the \(y\)-coordinate:
\[
f(1)=1^{2}-2(1)+3=1-2+3=2.
\]
Thus the vertex is at \( (1,2) \).

Step 4 – Find the y-intercept

Set \(x=0\):
\[
f(0)=0^{2}-2\cdot 0+3=3.
\]
The \(y\)-intercept is \( (0,3) \).

Step 5 – Check for x-intercepts

Compute the discriminant \(D=b^{2}-4ac\):
\[
D=(-2)^{2}-4(1)(3)=4-12=-8.
\]
Because the discriminant is negative, there are no real \(x\)-intercepts; the graph does not cross the \(x\)-axis.

Step 6 – Describe the graph

The graph is an upward-opening parabola with its minimum (vertex) at \( (1,2) \). It crosses the \(y\)-axis at \( (0,3) \) and lies entirely above the \(x\)-axis (no real roots).