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 (y = 0)

Set \( y = 0 \), so \( x^{2}-4x + 3=0 \). Factor the quadratic: \( (x - 1)(x - 3)=0 \). Thus, \( x=1 \) 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 \( (1,0) \), \( (3,0) \), \( (2,-1) \), \( (0,3) \), and \( (5,8) \).
3. Draw a parabola opening upward (since \( a = 1>0 \)) passing through these points, within the interval \( 0\leq x\leq5 \).

Answer:

The key points are \( (1,0) \) (x - intercept), \( (3,0) \) (x - intercept), \( (2,-1) \) (vertex/minimum), \( (0,3) \), and \( (5,8) \). The graph is a parabola opening upward passing through these points in the interval \( 0\leq x\leq5 \).