Question

1. consider the curve $y = x^{2}-6x + 2$.

12c using the minimum point on the curve, sketch a graph of the function.

Explanation:

Step1: Find the vertex (minimum point) of the parabola

For a quadratic function \(y = ax^{2}+bx + c\) (here \(a = 1\), \(b=-6\), \(c = 2\)), the \(x\) - coordinate of the vertex is given by \(x=-\frac{b}{2a}\).
Substitute \(a = 1\) and \(b=-6\) into the formula: \(x =-\frac{-6}{2\times1}=3\).
Then find the \(y\) - coordinate by substituting \(x = 3\) into \(y=x^{2}-6x + 2\): \(y=(3)^{2}-6\times3 + 2=9-18 + 2=-7\). So the vertex (minimum point) is \((3,-7)\).

Step2: Find the \(y\) - intercept

Set \(x = 0\) in \(y=x^{2}-6x + 2\). Then \(y=0^{2}-6\times0 + 2=2\). So the \(y\) - intercept is \((0,2)\).

Step3: Find the \(x\) - intercepts

Set \(y = 0\), so \(x^{2}-6x + 2=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=-6\), \(c = 2\).
\(x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times2}}{2\times1}=\frac{6\pm\sqrt{36 - 8}}{2}=\frac{6\pm\sqrt{28}}{2}=\frac{6\pm2\sqrt{7}}{2}=3\pm\sqrt{7}\approx3\pm2.65\). The \(x\) - intercepts are approximately \((0.35,0)\) and \((5.65,0)\).

Step4: Sketch the graph

• Plot the vertex \((3,-7)\), \(y\) - intercept \((0,2)\), and \(x\) - intercepts \((3-\sqrt{7},0)\) and \((3+\sqrt{7},0)\).
• Since \(a = 1>0\), the parabola opens upwards. Draw a smooth curve passing through these points.

Answer:

The graph is a parabola opening upwards with vertex at \((3,-7)\), \(y\) - intercept at \((0,2)\), and \(x\) - intercepts at approximately \((0.35,0)\) and \((5.65,0)\) (exact values \(x = 3\pm\sqrt{7}\)).