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# Help

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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?

Mar 28, 2018
edited by Guest  Mar 28, 2018

### Best Answer

#1
+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

Mar 28, 2018

### 1+0 Answers

#1
+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