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

To find where the graph intersects the x - axis, we set \(y = 0\). So we solve the equation \(x^{2}-4x + 3=0\). We can factor the quadratic as \((x - 1)(x - 3)=0\). Setting each factor equal to zero gives \(x-1 = 0\) 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)

For a quadratic function in the form \(y=ax^{2}+bx + c\), the x - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). For \(y=x^{2}-4x + 3\), \(a = 1\), \(b=-4\) and \(c = 3\). So \(x=-\frac{-4}{2\times1}=\frac{4}{2}=2\). To find the y - coordinate, 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

Let's choose \(x = 0\) and \(x = 5\) (within the interval \(0\leq x\leq5\)).

• 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 \((5,8)\).

Step4: Plot the points and draw the parabola

1. Draw the coordinate plane. Label the x - axis as "x - value" and the y - axis as "y - value".
2. Plot the points \((1,0)\), \((3,0)\), \((2,-1)\), \((0,3)\) and \((5,8)\).
3. Since the parabola opens upward (because \(a = 1>0\)), draw a smooth curve through these points.

To graph the function:

• X - intercepts: \((1,0)\) and \((3,0)\)
• Vertex (minimum): \((2,-1)\)
• Other points: \((0,3)\) and \((5,8)\)

Plot these points on a coordinate plane with x - axis labeled "x - value" and y - axis labeled "y - value" and draw a smooth upward - opening parabola through them.

Answer:

Step1: Find x - intercepts

To find where the graph intersects the x - axis, we set \(y = 0\). So we solve the equation \(x^{2}-4x + 3=0\). We can factor the quadratic as \((x - 1)(x - 3)=0\). Setting each factor equal to zero gives \(x-1 = 0\) 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)

For a quadratic function in the form \(y=ax^{2}+bx + c\), the x - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). For \(y=x^{2}-4x + 3\), \(a = 1\), \(b=-4\) and \(c = 3\). So \(x=-\frac{-4}{2\times1}=\frac{4}{2}=2\). To find the y - coordinate, 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

Let's choose \(x = 0\) and \(x = 5\) (within the interval \(0\leq x\leq5\)).

• 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 \((5,8)\).

Step4: Plot the points and draw the parabola

1. Draw the coordinate plane. Label the x - axis as "x - value" and the y - axis as "y - value".
2. Plot the points \((1,0)\), \((3,0)\), \((2,-1)\), \((0,3)\) and \((5,8)\).
3. Since the parabola opens upward (because \(a = 1>0\)), draw a smooth curve through these points.

To graph the function:

• X - intercepts: \((1,0)\) and \((3,0)\)
• Vertex (minimum): \((2,-1)\)
• Other points: \((0,3)\) and \((5,8)\)

Plot these points on a coordinate plane with x - axis labeled "x - value" and y - axis labeled "y - value" and draw a smooth upward - opening parabola through them.