question:\nconsider the quadratic function ( y = x^{2}-4x + 3 ) for ( 0leq xleq5 ).\ngraph the…

question:\nconsider the quadratic function ( y = x^{2}-4x + 3 ) for ( 0leq xleq5 ).\ngraph the function:\nplot the function ( y = x^{2}-4x + 3 ) on a coordinate plane. label the x - axis as \x - value\ and the y - axis as\n\y - value\. mark the coordinates of all key points where the graph intersects the x - axis, reaches its\nminimum (since the coefficient of ( x^{2} ) is positive, the parabola opens upward), and at least two other points\nthat help show the shape of the parabola within the given interval.

question:\nconsider the quadratic function ( y = x^{2}-4x + 3 ) for ( 0leq xleq5 ).\ngraph the function:\nplot the function ( y = x^{2}-4x + 3 ) on a coordinate plane. label the x - axis as \x - value\ and the y - axis as\n\y - value\. mark the coordinates of all key points where the graph intersects the x - axis, reaches its\nminimum (since the coefficient of ( x^{2} ) is positive, the parabola opens upward), and at least two other points\nthat help show the shape of the parabola within the given interval.

Answer

Explanation:

Step1: Find x - intercepts

Set (y = 0), then (x^{2}-4x + 3=0). Factor: ((x - 1)(x - 3)=0). So (x=1) or (x = 3). The points are ((1,0)) and ((3,0)).

Step2: Find the vertex

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

Answer:

Plot the points ((0,3)), ((1,0)), ((2,-1)), ((3,0)), ((5,8)) on the coordinate plane. Label the (x) - axis as "x - value" and the (y) - axis as "y - value". Connect these points with a smooth curve (a parabola opening upward) within the interval (0\leq x\leq5).