Question

complete the square to rewrite $y = x^2 - 6x + 5$ in vertex form. then state whether the vertex is a maximum or a minimum and give its coordinates. \\(\bigcirc\\) a. maximum at (3, -4) \\(\bigcirc\\) b. minimum at (3, -4) \\(\bigcirc\\) c. maximum at (-3, -4) \\(\bigcirc\\) d. minimum at (-3, -4)

Explanation:

Step1: Recall the vertex form and completing the square

The vertex form of a quadratic function is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. To complete the square for \(y=x^{2}-6x + 5\), we take the coefficient of \(x\), which is \(-6\), divide by \(2\) to get \(-3\), and then square it: \((-3)^{2}=9\).
We rewrite the equation as:
\(y=(x^{2}-6x + 9)-9 + 5\)

Step2: Simplify the equation

The expression \(x^{2}-6x + 9\) is a perfect square trinomial, which factors to \((x - 3)^{2}\). So we have:
\(y=(x - 3)^{2}-9 + 5=(x - 3)^{2}-4\)

Step3: Determine if the vertex is a maximum or minimum

For a quadratic function in the form \(y = ax^{2}+bx + c\) (or vertex form \(y=a(x - h)^{2}+k\)), if \(a>0\), the parabola opens upward, and the vertex is a minimum. In our function \(y=(x - 3)^{2}-4\), \(a = 1>0\), so the vertex is a minimum. The vertex coordinates from the vertex form \(y=(x - 3)^{2}-4\) are \((h,k)=(3,-4)\).

Answer:

B. Minimum at (3, -4)