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

Sagot :

MrRoyal

Answer:

[tex]y = (x - 3)^2 - 12[/tex]

[tex](3,-12)[/tex]

Step-by-step explanation:

Given

[tex]y = x^2 - 6x - 3[/tex]

Solving (a): In vertex form

The vertex form of an equation is:

[tex]y = a(x - h)^2 + k[/tex]

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

We have:

[tex]y = x^2 - 6x - 3[/tex]

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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

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[tex]y = x^2 - 6x - 3[/tex]

[tex]y = x^2 - 6x + 9 - 9 - 3[/tex]

[tex]y = x^2 - 6x + 9 - 12[/tex]

Factorize [tex]x^2 - 6x + 9[/tex]

[tex]y = x^2 - 3x-3x + 9 - 12[/tex]

[tex]y = x(x - 3)-3(x - 3) - 12[/tex]

Factor out x - 3

[tex]y = (x - 3)(x - 3) - 12[/tex]

Express as squares

[tex]y = (x - 3)^2 - 12[/tex]

Hence, the vertex form of [tex]y = x^2 - 6x - 3[/tex] is: [tex]y = (x - 3)^2 - 12[/tex]

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

In [tex]y = a(x - h)^2 + k[/tex]; the vertex is: (h,k)

The vertex of [tex]y = (x - 3)^2 - 12[/tex] will be [tex](3,-12)[/tex]

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