Question

sketch the graph of $f(x) = x^2 - 4x + 3$.

Explanation:

Step1: Find x-intercepts

Set $f(x)=0$, solve $x^2-4x+3=0$.
Factor: $(x-1)(x-3)=0$, so $x=1$ and $x=3$.

Step2: Find vertex (completing square)

Rewrite $f(x)$ in vertex form:

$$\begin{align*} f(x)&=x^2-4x+3\\ &=(x^2-4x+4)-4+3\\ &=(x-2)^2-1 \end{align*}$$

Vertex is at $(2, -1)$.

Step3: Find y-intercept

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

Step4: Determine parabola direction

Since coefficient of $x^2$ is $1>0$, parabola opens upward.

Answer:

The graph is an upward-opening parabola with:

• Vertex at $(2, -1)$
• x-intercepts at $(1, 0)$ and $(3, 0)$
• y-intercept at $(0, 3)$

(Plot these points and draw a smooth symmetric curve through them.)