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 and axis of symmetry

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex (and the axis of symmetry) is given by \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = -4 \), so:
\( x = -\frac{-4}{2(1)} = 2 \)
To find the y-coordinate of the vertex, substitute \( x = 2 \) into the function:
\( f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \)
So the vertex is \( (2, -1) \) and the axis of symmetry is \( x = 2 \).

Step2: Find the x-intercepts

Set \( f(x) = 0 \):
\( x^2 - 4x + 3 = 0 \)
Factor the quadratic:
\( (x - 1)(x - 3) = 0 \)
Set each factor equal to zero:
\( x - 1 = 0 \) or \( x - 3 = 0 \)
So \( x = 1 \) or \( x = 3 \). The x-intercepts are \( (1, 0) \) and \( (3, 0) \).

Step3: Find the y-intercept

Set \( x = 0 \) in the function:
\( f(0) = (0)^2 - 4(0) + 3 = 3 \)
So the y-intercept is \( (0, 3) \).

Step4: Sketch the graph

• Plot the vertex \( (2, -1) \), x-intercepts \( (1, 0) \) and \( (3, 0) \), y-intercept \( (0, 3) \).
• Draw the axis of symmetry \( x = 2 \).
• Since \( a = 1 > 0 \), the parabola opens upward. Connect the points smoothly to form the parabola.

Answer:

• Vertex: \( (2, -1) \)
• Axis of symmetry: \( x = 2 \)
• X-intercepts: \( (1, 0) \), \( (3, 0) \)
• Y-intercept: \( (0, 3) \)
• The graph is a parabola opening upward with the above - labeled points and axis of symmetry. (For the sketch, plot the points and draw the parabola as described in Step 4.)