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

question:\nconsider 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.

question:\nconsider 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), so (x^{2}-4x + 3=0). Factor: ((x - 1)(x - 3)=0). Solve: (x=1) or (x = 3). Points: ((1,0)) and ((3,0)).

Step2: Find vertex (minimum)

Use (x=-\frac{b}{2a}). Here (a = 1), (b=-4). So (x=\frac{4}{2}=2). Substitute (x = 2) into (y=x^{2}-4x + 3): (y=(2)^{2}-4(2)+3=4 - 8+3=-1). Vertex: ((2,-1)).

Step3: Find other points

For (x = 0): (y=(0)^{2}-4(0)+3=3). Point: ((0,3)). For (x = 5): (y=(5)^{2}-4(5)+3=25-20 + 3=8). Point: ((5,8)).

Answer:

Plot the points ((0,3)), ((1,0)), ((2,-1)), ((3,0)), ((5,8)) on a coordinate plane with (x) - axis labeled "x - value" and (y) - axis labeled "y - value" and draw a parabola through them.