To convert the quadratic function [tex]\( f(x) = 8x^2 + 4x \)[/tex] into its vertex form, we use the method of completing the square. The vertex form of a quadratic function is given by:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the quadratic function.
Let's start by completing the square:
1. Factor out the coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ f(x) = 8(x^2 + \frac{1}{2}x) \][/tex]
2. Complete the square inside the parentheses. In order to do that, we take half of the coefficient of [tex]\(x\)[/tex], square it, and add it inside the parentheses. Then, we subtract it outside.
[tex]\[ \text{Half of } \frac{1}{2} \text{ is } \frac{1}{4} \][/tex]
[tex]\[ \left(\frac{1}{4}\right)^2 = \frac{1}{16} \][/tex]
So, we add and subtract [tex]\(\frac{1}{16}\)[/tex] inside the parentheses:
[tex]\[ f(x) = 8 \left( x^2 + \frac{1}{2}x + \frac{1}{16} - \frac{1}{16} \right) \][/tex]
3. Rewrite the completed square and simplify:
[tex]\[ f(x) = 8 \left[ \left( x^2 + \frac{1}{2}x + \frac{1}{16} \right) - \frac{1}{16} \right] \][/tex]
[tex]\[ f(x) = 8 \left( x + \frac{1}{4} \right)^2 - 8 \cdot \frac{1}{16} \][/tex]
[tex]\[ f(x) = 8 \left( x + \frac{1}{4} \right)^2 - \frac{1}{2} \][/tex]
Thus, the function [tex]\( f(x) = 8x^2 + 4x \)[/tex] in vertex form is:
[tex]\[ f(x) = 8 \left( x + \frac{1}{4} \right)^2 - \frac{1}{2} \][/tex]
The correct option is:
[tex]\[ f(x) = 8\left(x+\frac{1}{4}\right)^2-\frac{1}{2} \][/tex]
