To express a function of the form [tex]f(x)= x^{2} +bx+c[/tex] in vertex form [tex]f(x)=a(x-h)^2+k[/tex] where (h,k) is the vertex of the parabola, we need to find the vertex first.
To find the vertex we are going to use the vertex formula: [tex]h= \frac{-b}{2a} [/tex], and [tex]k[/tex] will be the function evaluated at [tex]h[/tex].
We can infer from our function that [tex]a=1[/tex] and [tex]b=-2[/tex]. So lets find [tex]h[/tex]:
[tex]h= \frac{-b}{2a} [/tex]
[tex]h= \frac{-(-2)}{2(1)} [/tex]
[tex]h= \frac{2}{2} [/tex]
[tex]h=1[/tex]
Now that we have [tex]h[/tex], we can evaluate the function at 1 to find [tex]k[/tex]:
[tex]f(x)=4+ x^{2} -2x[/tex]
[tex]f(1)=4+(1)^2-2(1)[/tex]
[tex]f(1)=4+1-2[/tex]
[tex]f(1)=3[/tex]
We have the vertex (1,3) of our parabola, so we can use its vertex form:
[tex]f(x)=a(x-h)^2+k[/tex]
[tex]f(x)=1(x-1)^2+3[/tex]
[tex]f(x)=(x-1)^2+3[/tex]
We can conclude that the vertex for of our parabola is f(x) = (x + 1)2 + 3.
