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 x-intercepts (set \( f(x) = 0 \))

\( x^2 - 4x + 3 = 0 \)
Factor: \( (x - 1)(x - 3) = 0 \)
Solutions: \( x = 1 \), \( x = 3 \). So x-intercepts: \( (1, 0) \), \( (3, 0) \).

Step2: Find y-intercept (set \( x = 0 \))

\( f(0) = 0^2 - 4(0) + 3 = 3 \). So y-intercept: \( (0, 3) \).

Step3: Find vertex and axis of symmetry

For \( f(x) = ax^2 + bx + c \), axis of symmetry: \( x = -\frac{b}{2a} \).
Here, \( a = 1 \), \( b = -4 \), so \( x = -\frac{-4}{2(1)} = 2 \).
Vertex x-coordinate: \( x = 2 \).
Vertex y-coordinate: \( f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \).
Vertex: \( (2, -1) \), axis of symmetry: \( x = 2 \).

Step4: Sketch the parabola

• Opens upward (since \( a = 1 > 0 \)).
• Plot intercepts, vertex, and draw the parabola symmetric about \( x = 2 \).

Answer:

• x-intercepts: \( (1, 0) \), \( (3, 0) \)
• y-intercept: \( (0, 3) \)
• Vertex: \( (2, -1) \)
• Axis of symmetry: \( x = 2 \)
• The graph is a parabola opening upward with the above - labeled features.