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

Complete the square for \( f(x) = x^2 - 4x + 3 \).
\( f(x) = x^2 - 4x + 4 - 4 + 3 = (x - 2)^2 - 1 \).
So the vertex form is \( f(x) = (x - 2)^2 - 1 \), vertex is \( (2, -1) \).

Step2: Find x-intercepts

Set \( f(x) = 0 \): \( x^2 - 4x + 3 = 0 \).
Factor: \( (x - 1)(x - 3) = 0 \).
Solutions: \( x = 1 \) and \( x = 3 \), so x-intercepts are \( (1, 0) \) and \( (3, 0) \).

Step3: Find y-intercept

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

Step4: Find axis of symmetry

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

Step5: Sketch the graph

• The parabola opens upward (since \( a = 1 > 0 \)).
• Plot vertex \( (2, -1) \), x-intercepts \( (1, 0) \), \( (3, 0) \), y-intercept \( (0, 3) \).
• Draw the parabola symmetric about \( x = 2 \).

Answer:

Vertex: \((2, -1)\), x-intercepts: \((1, 0)\), \((3, 0)\), y-intercept: \((0, 3)\), axis of symmetry: \(x = 2\) (graph is a upward - opening parabola passing through these points, symmetric about \(x = 2\))