Question

Put the quadratic into vertex form and state the coordinates of the vertex. y= y= \,\,x^2-6x-3 x 2 −6x−3

Answer 1

Answer:

$y = (x - 3)^2 - 12$

$(3,-12)$

Step-by-step explanation:

Given

$y = x^2 - 6x - 3$

Solving (a): In vertex form

The vertex form of an equation is:

$y = a(x - h)^2 + k$

To do this, we make use of completing the square method.

We have:

$y = x^2 - 6x - 3$

------------------------------------------------------------------

Take the coefficient of x (i.e. -6)

Divide by 2; -6/2 = -3

Square it: (-3)^2 = 9

Add and subtract the result to the equation

------------------------------------------------------------------

$y = x^2 - 6x - 3$

$y = x^2 - 6x + 9 - 9 - 3$

$y = x^2 - 6x + 9 - 12$

Factorize $x^2 - 6x + 9$

$y = x^2 - 3x-3x + 9 - 12$

$y = x(x - 3)-3(x - 3) - 12$

Factor out x - 3

$y = (x - 3)(x - 3) - 12$

Express as squares

$y = (x - 3)^2 - 12$

Hence, the vertex form of $y = x^2 - 6x - 3$ is: $y = (x - 3)^2 - 12$

Solving (b): State the coordinates of the vertex.

In $y = a(x - h)^2 + k$; the vertex is: (h,k)

The vertex of $y = (x - 3)^2 - 12$ will be $(3,-12)$