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 equation of the axis of symmetry) is given by \( x=-\frac{b}{2a} \). For \( f(x)=x^{2}-4x + 3 \), \( a = 1 \), \( b=-4 \), \( c = 3 \). So \( x=-\frac{-4}{2\times1}=2 \). To find the y - coordinate of the vertex, substitute \( x = 2 \) into the function: \( f(2)=2^{2}-4\times2 + 3=4 - 8+3=-1 \). So the vertex is \( (2,-1) \) and the axis of symmetry is \( x = 2 \).

Step2: Find the x - intercepts

Set \( f(x)=0 \), so \( x^{2}-4x + 3=0 \). Factor the quadratic: \( (x - 1)(x - 3)=0 \). Setting each factor equal to zero gives \( x - 1=0\) 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: \( f(0)=0^{2}-4\times0+3 = 3 \). So the y - intercept is \( (0,3) \).

Step4: Sketch the graph

• Plot the vertex \( (2,-1) \), the x - intercepts \( (1,0) \), \( (3,0) \), and the y - intercept \( (0,3) \).
• Draw a parabola (since \( a = 1>0 \), it opens upwards) symmetric about the line \( x = 2 \) passing through these points.

Answer:

The vertex is \( (2,-1) \), the axis of symmetry is \( x = 2 \), the x - intercepts are \( (1,0) \) and \( (3,0) \), the y - intercept is \( (3,0) \) (wait, correction: y - intercept is \( (0,3) \)), and the graph is a parabola opening upwards with the above - mentioned key points. (The sketching process is as described in the steps above, with the parabola passing through \( (0,3) \), \( (1,0) \), \( (2,-1) \), \( (3,0) \) and symmetric about \( x = 2 \))