Answer:
Table B
Vertex is (3,-5)
Step-by-step explanation:
We are given with an equation of a parabola [tex]y=x^2-6x+4[/tex]
Let is convert it into standard form of a parabola
[tex]y=x^2-6x+4[/tex]
adding and subtracting 9 in the right hand side of the =
[tex]y=x^2-6x+9-9+4[/tex]
[tex]y=x^2-2\times 3\times x+ 3^2-9+4[/tex]
the first three terms of the right hand side forms the expression of square of difference
[tex]a^2-2 \times a \times b+b^2 = (a-b)^2[/tex]
Hence
[tex]y=(x-3)^2-5[/tex]
adding 5 on both sides we get
[tex](y+5)=(x-3)^2[/tex]
Comparing it with the standard equation of a parabola
[tex]X^2=4\times \frac{1}{4} \times Y[/tex]
where [tex]X=x-3[/tex] and [tex]Y=y+5[/tex]
The vertex of [tex]X^2=4\times \frac{1}{4} \times Y[/tex] will be (0,0)
and thus vertex of
[tex](y+5)=(x-3)^2[/tex] will be (3,-5)
Hence the Table B is our right answer