sketch the graph of the quadratic function ( f(x)=x^{2}-4x + 3 ), labeling its vertex, x - intercepts, y…

Explanation:

Step1: Find y-intercept

Set $x=0$: $f(0)=0^2-4(0)+3=3$.
Y-intercept: $(0,3)$

Step2: Find x-intercepts

Set $f(x)=0$: $x^2-4x+3=0$. Factor: $(x-1)(x-3)=0$.
X-intercepts: $(1,0),(3,0)$

Step3: Find axis of symmetry

Formula $x=-\frac{b}{2a}$: $a=1,b=-4$. $x=-\frac{-4}{2(1)}=2$.
Axis of symmetry: $x=2$

Step4: Find vertex

Substitute $x=2$ into $f(x)$: $f(2)=2^2-4(2)+3=-1$.
Vertex: $(2,-1)$

Step5: Sketch the graph

Plot intercepts, vertex; draw parabola opening up (since $a>0$), symmetric about $x=2$.

Answer:

• Vertex: $(2, -1)$
• X-intercepts: $(1, 0)$ and $(3, 0)$
• Y-intercept: $(0, 3)$
• Axis of symmetry: $x = 2$
  (Graph: Parabola opening upward passing through these points, symmetric over $x=2$)