Question

sketch the graph of the quadratic function $f(x) = x^2 - 4x + 3$, labeling its vertex, x-intercepts, y-intercept, and axis of symmetry.

Explanation:

Step1: Find the vertex form (completing the square)

To find the vertex form of \( f(x) = x^2 - 4x + 3 \), we complete the square.
The coefficient of \( x \) is -4, half of -4 is -2, and squaring -2 gives 4. So we add and subtract 4:
\[

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

\]
From the vertex form \( f(x) = a(x - h)^2 + k \), the vertex is \( (h, k) \). So the vertex of \( f(x) \) is \( (2, -1) \).

Step2: Find the axis of symmetry

For a quadratic function in the form \( f(x) = a(x - h)^2 + k \), the axis of symmetry is the vertical line \( x = h \). From the vertex form we found, \( h = 2 \), so the axis of symmetry is \( x = 2 \).

Step3: Find the x-intercepts

To find the x-intercepts, we set \( f(x) = 0 \):
\[
x^2 - 4x + 3 = 0
\]
Factor the quadratic equation:
\[
(x - 1)(x - 3) = 0
\]
Setting each factor equal to zero gives \( x - 1 = 0 \) or \( x - 3 = 0 \), so the x-intercepts are \( x = 1 \) and \( x = 3 \), which correspond to the points \( (1, 0) \) and \( (3, 0) \).

Step4: Find the y-intercept

To find the y-intercept, we set \( x = 0 \) in the function:
\[
f(0) = 0^2 - 4(0) + 3 = 3
\]
So the y-intercept is \( (0, 3) \).

Step5: Sketch the graph

• The parabola opens upward because the coefficient of \( x^2 \) (which is 1) is positive.
• Plot the vertex \( (2, -1) \), the x-intercepts \( (1, 0) \) and \( (3, 0) \), the y-intercept \( (0, 3) \), and draw the axis of symmetry \( x = 2 \). Then sketch the parabola symmetric about the axis of symmetry, passing through these points.

Answer:

• Vertex: \( (2, -1) \)
• Axis of Symmetry: \( x = 2 \)
• x-intercepts: \( (1, 0) \) and \( (3, 0) \)
• y-intercept: \( (0, 3) \)
• The graph is a parabola opening upward with the above - labeled features. (To sketch, plot the points and draw the parabola symmetric about \( x = 2 \).)