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 the quadratic: \((x - 1)(x - 3)=0\). 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 two 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:

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

Answer:

• x - intercepts: \((1,0)\), \((3,0)\)
• Minimum point (vertex): \((2,-1)\)
• Other points: \((0,3)\), \((5,8)\) (and the parabola is drawn through these points with x - axis labeled "x - value" and y - axis labeled "y - value")