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Here is an equation of a parabola:

 

y = 1/6 x^2 + 2x + 11

 

What are the coordinates of the vertex of the parabola?

Guest Mar 28, 2018
edited by Guest  Mar 28, 2018

Best Answer 

 #1
avatar+7155 
+3

\(y=\frac16x^2+2x+11\)

                                                Multiply through by  6 .

\(6y=x^2+12x+66\)

                                                Subtract  66  from both sides of the equation.

\(6y-66=x^2+12x\)

                                                Add  36  to both sides to complete the square on the right side.

\(6y-30=x^2+12x+36\)

                                                Factor the right side as a perfect square trinomial.

\(6y-30=(x+6)^2\)

                                                Factor  6 out of the terms on the left side.

\(6(y-5)=(x+6)^2\)

 

Now that the equation is in this form we can see that the vertex is  (-6, 5) .

 

Here's a graph to check it  https://www.desmos.com/calculator/wrvhzfuag3

hectictar  Mar 28, 2018
 #1
avatar+7155 
+3
Best Answer

\(y=\frac16x^2+2x+11\)

                                                Multiply through by  6 .

\(6y=x^2+12x+66\)

                                                Subtract  66  from both sides of the equation.

\(6y-66=x^2+12x\)

                                                Add  36  to both sides to complete the square on the right side.

\(6y-30=x^2+12x+36\)

                                                Factor the right side as a perfect square trinomial.

\(6y-30=(x+6)^2\)

                                                Factor  6 out of the terms on the left side.

\(6(y-5)=(x+6)^2\)

 

Now that the equation is in this form we can see that the vertex is  (-6, 5) .

 

Here's a graph to check it  https://www.desmos.com/calculator/wrvhzfuag3

hectictar  Mar 28, 2018

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