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 the x - intercepts

Set \(y = 0\), then \(x^{2}-4x + 3=0\). 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

The formula for the x - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). Here \(a = 1\), \(b=-4\), so \(x=\frac{4}{2}=2\). Substitute \(x = 2\) into the function: \(y=(2)^{2}-4\times(2)+3=4-8 + 3=-1\). 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 a coordinate plane. Connect these points with a smooth curve (a parabola opening upward) to graph the function \(y=x^{2}-4x + 3\) for \(0\leq x\leq5\).