Here is an equation of a parabola:
y = 1/6 x^2 + 2x + 11
What are the coordinates of the vertex of the parabola?
+9466 \(y=\frac16x^2+2x+11\)
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
+9466 \(y=\frac16x^2+2x+11\)
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
