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Need help with this maybe question

Need help with this maybe question-example-1

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Answer:

the vertex is:

(2, -1)

Explanation:

First solve the equation for the variable y


x^2-16y-4x-12=0

Add 16y on both sides of the equation


16y=x^2-16y+16y-4x-12


16y=x^2-4x-12

Notice that now the equation has the general form of a parabola


ax^2 +bx +c

In this case


a=1\\b=-4\\c=-12

Add
((b)/(2)) ^ 2 and subtract
((b)/(2)) ^ 2 on the right side of the equation


((b)/(2)) ^ 2=((-4)/(2)) ^ 2


((b)/(2)) ^ 2=(-2) ^ 2


((b)/(2)) ^ 2=4


16y=(x^2-4x+4)-4-12

Factor the expression that is inside the parentheses


16y=(x-2)^2-16

Divide both sides of the equality between 16


(16)/(16)y=(1)/(16)(x-2)^2-(16)/(16)


y=(1)/(16)(x-2)^2-1

For an equation of the form


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

the vertex is: (h, k)

In this case


h=2\\k =-1

the vertex is:

(2, -1)

User SamPassmore
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