Answer:
Vertex form:
[tex]f(x)=8(x+\frac{1}{4})^2-\frac{1}{2}[/tex]
Step-by-step explanation:
Given: [tex]f(x)=8x^2+4x[/tex]
We need to write in vertex form.
Vertex form:
[tex]y=a(x-h)^2+k[/tex]
vertex: (h,k)
[tex]f(x)=8x^2+4x[/tex]
Step 1: Take out 8 common from each term
[tex]f(x)=8(x^2+\frac{1}{2}x)[/tex]
Step 2: Add and subtract square of half of coefficient of x
[tex]f(x)=8(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16})[/tex]
Step 3: Factor the term inside parentheses
[tex]f(x)=8(x^2+2\cdot \frac{1}{4}\cdot x+(\frac{1}{4})^2)-8\cdot \frac{1}{16})[/tex]
[tex]f(x)=8(x+\frac{1}{4})^2-\frac{1}{2}[/tex] [tex]\because (a^2+2ab+b^2)=(a+b)^2[/tex]
Hence, The vertex form of f(x)
[tex]f(x)=8(x+\frac{1}{4})^2-\frac{1}{2}[/tex]
