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

Step2: Find the x - intercepts

Set \( f(x)=0 \), so \( x^{2}-4x + 3=0 \).
Factor the quadratic equation: \( (x - 1)(x - 3)=0 \).
Set each factor equal to zero: \( x - 1=0 \) or \( x - 3=0 \).
Solving these, we get \( x = 1 \) or \( x = 3 \). So 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

• The parabola opens upward because \( a = 1>0 \).
• Plot the vertex \( (2,-1) \), the axis of symmetry \( x = 2 \), the x - intercepts \( (1,0) \), \( (3,0) \) and the y - intercept \( (0,3) \). Then draw a smooth parabola passing through these points.

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 - mentioned features. (To sketch the graph, plot the points and draw a smooth curve through them.)