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

Explanation:

Step1: Find the axis of symmetry

For a quadratic function $y = ax^{2}+bx + c$, the axis of symmetry is given by $x=-\frac{b}{2a}$. Here $a = 1$, $b=-4$, so $x =-\frac{-4}{2\times1}=2$.

Step2: Find the vertex

Substitute $x = 2$ into the function $f(x)=x^{2}-4x + 3$. Then $f(2)=2^{2}-4\times2 + 3=4 - 8+3=-1$. So the vertex is $(2,-1)$.

Step3: Find the x - intercepts

Set $y = f(x)=0$, so $x^{2}-4x + 3=0$. Factor the quadratic equation: $(x - 1)(x - 3)=0$. Then $x=1$ or $x = 3$. The x - intercepts are $(1,0)$ and $(3,0)$.

Step4: Find the y - intercept

Set $x = 0$, then $f(0)=0^{2}-4\times0+3=3$. The y - intercept is $(0,3)$.

Answer:

To sketch the graph: Draw a parabola opening upwards (since $a=1>0$) with axis of symmetry $x = 2$, vertex at $(2,-1)$, x - intercepts at $(1,0)$ and $(3,0)$, and y - intercept at $(0,3)$.