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 \( f(x)=x^{2}-4x + 3 \), \( a = 1 \), \( b=-4 \), \( c = 3 \).
\( x=-\frac{-4}{2\times1}=\frac{4}{2}=2 \)
To find the y - coordinate of the vertex, substitute \( x = 2 \) into the function:
\( f(2)=2^{2}-4\times2 + 3=4-8 + 3=-1 \)
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 equation: \( x^{2}-4x + 3=(x - 1)(x - 3)=0 \)
Set each factor equal to zero: \( x-1=0\Rightarrow x = 1 \); \( x - 3=0\Rightarrow x = 3 \)
So the x - intercepts are \( (1,0) \) and \( (3,0) \).

Step3: Find the y - intercept

Set \( x = 0 \) in the function \( f(x)=x^{2}-4x + 3 \).
\( f(0)=0^{2}-4\times0+3 = 3 \)
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 x - intercepts \( (1,0) \) and \( (3,0) \), the y - intercept \( (0,3) \).
• Draw a parabola symmetric about the line \( x = 2 \) 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.