Question

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 $y=0$: $x^2-4x+3=0 \Rightarrow (x-1)(x-3)=0 \Rightarrow x=1,3$. Points: $(1,0),(3,0)$.

Step2: Find vertex (minimum)

Vertex x-coordinate: $x=-\frac{b}{2a}=-\frac{-4}{2(1)}=2$. Substitute $x=2$: $y=2^2-4(2)+3=-1$. Point: $(2,-1)$.

Step3: Find boundary points

At $x=0$: $y=0-0+3=3$ (point $(0,3)$). At $x=5$: $y=25-20+3=8$ (point $(5,8)$).

Step4: Plot points and graph

Plot $(0,3),(1,0),(2,-1),(3,0),(5,8)$; draw upward-opening parabola through them. Label axes.

Answer:

The graph is an upward-opening parabola on $0\leq x\leq5$ with key points: $(0,3)$, $(1,0)$, $(2,-1)$, $(3,0)$, $(5,8)$. Axes labeled "x-value" (x-axis) and "y-value" (y-axis).