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 (y=0)

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

Step2: Find vertex (minimum point)

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

Step3: Find other points

Choose \( x = 0 \): \( y = 0 - 0 + 3 = 3 \), point \( (0, 3) \).
Choose \( x = 5 \): \( y = 25 - 20 + 3 = 8 \), point \( (5, 8) \).

Step4: Plotting

• Draw x-axis (label "x - value") and y-axis (label "y - value").
• Plot \( (1, 0) \), \( (3, 0) \), \( (2, -1) \), \( (0, 3) \), \( (5, 8) \).
• Draw a parabola opening upward through these points.

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".