Answer:
The vertex of the function is (2, -3).
Step-by-step explanation:
Given:
[tex]y=x^{2}-4x+1[/tex]
So, to find the vertex of the function we will get the equation in the form:
[tex]y=ax^{2} +bx+c[/tex]
[tex]y=1x^{2}+(-4)x+1[/tex]
So, [tex]a=1,b=-4,c=1[/tex]
Then, we calculate the x-coordinate of the vertex:
[tex]x=\frac{-b}{2a}[/tex]
[tex]x=\frac{-(-4)}{2\times1}\\x=\frac{4}{2}[/tex]
[tex]x=2[/tex]
And now, we get the [tex]y[/tex] value of vertex of the function:
[tex]y=1x^{2}-4x+1[/tex]
[tex]y=1\times 2^{2}+(-4)\times (2)+1[/tex]
[tex]y=1\times 4-8+1[/tex] (when the opposite signs multiply the result is negative)
[tex]y=4-8+1[/tex]
[tex]y=-3[/tex]
Therefore, the vertex is at [tex](x,y)=(2,-3)[/tex].
