Q. Vertex of \( g(x) = 8x^2 – 48x + 65 \)

Find the vertex of the quadratic function

Given the quadratic function \(g(x)=8x^{2}-48x+65\), we will find its vertex by completing the square. Each step is explained in detail.

1. Identify coefficients: For a quadratic in the form \(ax^{2}+bx+c\), here

   \(a=8,\quad b=-48,\quad c=65\).

2. Factor out \(a\) from the first two terms: Factor 8 from \(8x^{2}-48x\) to prepare for completing the square.

   \(g(x)=8\bigl(x^{2}-6x\bigr)+65\).

   Explanation: dividing the first two terms by 8 gives \(x^{2}-6x\); the constant 65 remains outside for now.

3. Complete the square inside the parentheses: For the quadratic \(x^{2}-6x\), take half of the coefficient of \(x\), which is \(-6\). Half of \(-6\) is \(-3\). Square that to get 9. Add and subtract 9 inside the parentheses so the expression becomes a perfect square plus a correction term.

   \(x^{2}-6x = \bigl(x^{2}-6x+9\bigr)-9 = (x-3)^{2}-9\).

   Explanation: \((x-3)^{2}=x^{2}-6x+9\), so adding and subtracting 9 does not change the value but allows factoring.

4. Substitute back and simplify: Replace the completed-square expression into \(g(x)\).

   \(g(x)=8\bigl((x-3)^{2}-9\bigr)+65\).

   Distribute the 8 and combine constants:

   \(g(x)=8(x-3)^{2}-72+65\).

   \(g(x)=8(x-3)^{2}-7\).

   Explanation: Multiplying \(-9\) by 8 gives \(-72\); adding 65 yields \(-7\).

5. Read off the vertex from the vertex form: The vertex form is \(g(x)=a(x-h)^{2}+k\), where the vertex is \((h,k)\). Here the function is \(g(x)=8(x-3)^{2}-7\), so

   Vertex: \((h,k)=(3,-7)\).

   Explanation: The expression \((x-3)^{2}\) indicates \(h=3\); the constant term outside is \(k=-7\).

Final answer: The vertex of \(g(x)=8x^{2}-48x+65\) is \((3,-7)\).