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 the quadratic: \(x^{2}-4x + 3=(x - 1)(x - 3)=0\). So \(x=1\) or \(x = 3\). The x - intercepts are \((1,0)\) and \((3,0)\).

Step2: Find the vertex (minimum point)

For a quadratic \(y=ax^{2}+bx + c\), the x - coordinate of the vertex is \(x=-\frac{b}{2a}\). Here, \(a = 1\), \(b=-4\), so \(x=-\frac{-4}{2\times1}=2\). Substitute \(x = 2\) into the function: \(y=(2)^{2}-4\times2 + 3=4-8 + 3=-1\). The vertex (minimum point) is \((2,-1)\).

Step3: Find other points

• When \(x = 0\): \(y=0^{2}-4\times0+3 = 3\), so the point is \((0,3)\).
• When \(x = 5\): \(y=5^{2}-4\times5+3=25 - 20 + 3=8\), so the point is \((5,8)\).

To graph the function:

1. Draw the coordinate plane. Label the x - axis as "x - value" and the y - axis as "y - value".
2. Plot the points: \((0,3)\), \((1,0)\), \((2,-1)\), \((3,0)\), \((5,8)\).
3. Draw a smooth parabola (opening upward) passing through these points, restricted to the interval \(0\leq x\leq5\).

Answer:

The key points are \((0,3)\), \((1,0)\), \((2,-1)\), \((3,0)\), \((5,8)\). The graph is a parabola opening upward passing through these points in the interval \(0\leq x\leq5\).