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)

To find where the graph intersects the x - axis, we solve the equation \(x^{2}-4x + 3=0\). We can factor the quadratic equation as \((x - 1)(x - 3)=0\). Setting each factor equal to zero gives us \(x-1 = 0\) or \(x - 3=0\), so \(x = 1\) and \(x=3\). The x - intercepts are \((1,0)\) and \((3,0)\).

Step2: Find the vertex (minimum point)

The x - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). For the function \(y=x^{2}-4x + 3\), \(a = 1\), \(b=-4\), and \(c = 3\). So \(x=-\frac{-4}{2\times1}=\frac{4}{2}=2\). To find the y - coordinate of the vertex, we substitute \(x = 2\) into the function: \(y=(2)^{2}-4\times(2)+3=4 - 8 + 3=-1\). So the vertex (minimum point) is \((2,-1)\).

Step3: Find additional points

Let's choose some x - values within the interval \(0\leq x\leq5\) and find the corresponding y - values.

• When \(x = 0\): \(y=(0)^{2}-4\times(0)+3=3\). So the point is \((0,3)\).
• When \(x = 4\): \(y=(4)^{2}-4\times(4)+3=16-16 + 3=3\). So the point is \((4,3)\).
• When \(x = 5\): \(y=(5)^{2}-4\times(5)+3=25-20 + 3=8\). So the point is \((5,8)\).

Graphing Steps:

1. Draw the coordinate plane: Label the x - axis as "x - value" and the y - axis as "y - value".
2. Plot the key points:

• Plot the x - intercepts \((1,0)\) and \((3,0)\).
• Plot the vertex \((2,-1)\).
• Plot the additional points \((0,3)\), \((4,3)\), and \((5,8)\).

1. Draw the parabola: Since the coefficient of \(x^{2}\) is positive, the parabola opens upward. Connect the points with a smooth curve.

(Note: Since this is a text - based response, we can't actually draw the graph here, but the above steps describe how to plot the function \(y=x^{2}-4x + 3\) for \(0\leq x\leq5\).)

Answer:

Step1: Find x - intercepts (set y = 0)

To find where the graph intersects the x - axis, we solve the equation \(x^{2}-4x + 3=0\). We can factor the quadratic equation as \((x - 1)(x - 3)=0\). Setting each factor equal to zero gives us \(x-1 = 0\) or \(x - 3=0\), so \(x = 1\) and \(x=3\). The x - intercepts are \((1,0)\) and \((3,0)\).

Step2: Find the vertex (minimum point)

The x - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). For the function \(y=x^{2}-4x + 3\), \(a = 1\), \(b=-4\), and \(c = 3\). So \(x=-\frac{-4}{2\times1}=\frac{4}{2}=2\). To find the y - coordinate of the vertex, we substitute \(x = 2\) into the function: \(y=(2)^{2}-4\times(2)+3=4 - 8 + 3=-1\). So the vertex (minimum point) is \((2,-1)\).

Step3: Find additional points

Let's choose some x - values within the interval \(0\leq x\leq5\) and find the corresponding y - values.

• When \(x = 0\): \(y=(0)^{2}-4\times(0)+3=3\). So the point is \((0,3)\).
• When \(x = 4\): \(y=(4)^{2}-4\times(4)+3=16-16 + 3=3\). So the point is \((4,3)\).
• When \(x = 5\): \(y=(5)^{2}-4\times(5)+3=25-20 + 3=8\). So the point is \((5,8)\).

Graphing Steps:

1. Draw the coordinate plane: Label the x - axis as "x - value" and the y - axis as "y - value".
2. Plot the key points:

• Plot the x - intercepts \((1,0)\) and \((3,0)\).
• Plot the vertex \((2,-1)\).
• Plot the additional points \((0,3)\), \((4,3)\), and \((5,8)\).

1. Draw the parabola: Since the coefficient of \(x^{2}\) is positive, the parabola opens upward. Connect the points with a smooth curve.

(Note: Since this is a text - based response, we can't actually draw the graph here, but the above steps describe how to plot the function \(y=x^{2}-4x + 3\) for \(0\leq x\leq5\).)