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

consider the quadratic function $y = x^2 - 4x + 3$ for $0 \\leq x \\leq 5$.\n\ngraph the function:\n\nplot 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.
Answer
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:
- Draw the coordinate plane, label x - axis as "x - value" and y - axis as "y - value".
- Plot the points ((1,0)), ((3,0)), ((2,-1)), ((0,3)), ((5,8)).
- 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")