Question

question:
consider the quadratic function $y = x^2 - 4x + 3$ for $0 \leq x \leq 5$.

graph the function:

plot the function $y = x^2 - 4x + 3$ on a coordinate plane. label the x - axis as \x - value\ and the y - axis as \y - value\. mark the coordinates of all key points where the graph intersects the x - axis, reaches its minimum (since the coefficient of $x^2$ is positive, the parabola opens upward), and at least two other points that help show the shape of the parabola within the given interval.

Explanation:

Step1: Find x-intercepts

Set \( y = 0 \), solve \( x^{2}-4x + 3=0 \). Factor: \( (x - 1)(x - 3)=0 \), so \( x = 1 \) or \( x = 3 \). Intercepts: \( (1,0) \), \( (3,0) \).

Step2: Find vertex (minimum)

For \( y=ax^{2}+bx + c \), vertex \( x=-\frac{b}{2a} \). Here \( a = 1 \), \( b=-4 \), so \( x = 2 \). Substitute \( x = 2 \) into \( y \): \( y=2^{2}-4\times2 + 3=-1 \). Vertex: \( (2,-1) \).

Step3: Find other points

At \( x = 0 \): \( y=0^{2}-4\times0 + 3 = 3 \), point \( (0,3) \). At \( x = 5 \): \( y=5^{2}-4\times5 + 3 = 8 \), point \( (5,8) \).

Step4: Graph

• Draw coordinate plane, label axes.
• Plot \( (1,0) \), \( (3,0) \), \( (2,-1) \), \( (0,3) \), \( (5,8) \).
• Draw parabola through these points, opening upward.

Answer:

The graph is a parabola opening upward with x - intercepts at \((1,0)\) and \((3,0)\), vertex (minimum) at \((2, - 1)\), and additional points \((0,3)\) and \((5,8)\) plotted on a coordinate plane with x - axis labeled "x - value" and y - axis labeled "y - value".