Answer:
The function in vertex form is [tex]f(x)=(x-(-\frac{1}{2}))^2+\frac{3}{4}[/tex]
Step-by-step explanation:
Given: Function [tex]f(x)=x^2+x+1[/tex]
We have to write the given function [tex]f(x)=x^2+x+1[/tex] in vertex form.
Consider the given function [tex]f(x)=x^2+x+1[/tex]
Using algebraic identity [tex](a+b)^2=a^2+b^2+2ab[/tex]
Comparing a = x , 2ab = x
Also, 2b = 1 ⇒[tex] b=\frac{1}{2}[/tex]
Add and subtract [tex]b^2=\frac{1}{4}[/tex] , we get,
[tex]f(x)=x^2+x+1+\frac{1}{4}-\frac{1}{4}[/tex]
Simplify, we have,
[tex]f(x)=(x+\frac{1}{2})^21-\frac{1}{4}[/tex]
Simplify, we have,
[tex]f(x)=(x+\frac{1}{2})^2+\frac{3}{4}[/tex]
Thus, The function in vertex form is [tex]f(x)=(x-(-\frac{1}{2}))^2+\frac{3}{4}[/tex]
