Question

Which table contains the vertex of the graph represented by y=x^2-6x+4?

Answer 1

Answer:

Table B

Vertex is (3,-5)

Step-by-step explanation:

We are given with an equation of a parabola $y=x^2-6x+4$

Let is convert it into standard form of a parabola

$y=x^2-6x+4$

adding and subtracting 9 in the right hand side of the =

$y=x^2-6x+9-9+4$

$y=x^2-2\times 3\times x+ 3^2-9+4$

the first three terms of the right hand side forms the expression of square of difference

$a^2-2 \times a \times b+b^2 = (a-b)^2$

Hence

$y=(x-3)^2-5$

adding 5 on both sides we get

$(y+5)=(x-3)^2$

Comparing it with the standard equation of a parabola

$X^2=4\times \frac{1}{4} \times Y$

where $X=x-3$ and $Y=y+5$

The vertex of $X^2=4\times \frac{1}{4} \times Y$ will be (0,0)

and thus vertex of

$(y+5)=(x-3)^2$ will be (3,-5)

Hence the Table B is our right answer