# How do you find the vertex of (x + 6)^2 = -36(y − 3)?

Jan 30, 2016

vertex$= \left(- 6 , 3\right)$

#### Explanation:

1. Expand both sides of the equation.

${\left(x + 6\right)}^{2} = - 36 \left(y - 3\right)$

${x}^{2} + 12 x + 36 = - 36 y + 108$

2. Isolate for y.
Recall that general equation for a quadratic equation in standard form is $y = a {x}^{2} + b x + c$. Thus, isolate for $y$.

${x}^{2} + 12 x + 36 - 108 = - 36 y$

${x}^{2} + 12 x - 72 = - 36 y$

$y = - \frac{1}{36} {x}^{2} - \frac{1}{3} x + 2$

3. Factor -1/36 from the first two terms.
To find the vertex, we must complete the square. We can do this by first factoring $- \frac{1}{36}$ from the first two terms.

$y = - \frac{1}{36} \left({x}^{2} + 12 x\right) + 2$

4. Rewrite the bracketed terms as a perfect square trinomial.
The value of $c$ in a perfect square trinomial is ${\left(\frac{b}{2}\right)}^{2}$. Thus, divide $12$ by $2$ and square the value.

$y = - \frac{1}{36} \left({x}^{2} + 12 x + {\left(\frac{12}{2}\right)}^{2}\right) + 2$

$y = - \frac{1}{36} \left({x}^{2} + 12 x + 36\right) + 2$

5. Subtract 36 from the perfect square trinomial.
We cannot just add $36$ to the perfect square trinomial, so we must subtract $36$ from the $36$ we just added.

y=-1/36(x^2+12x+36 color(red)(-36))+2

6. Multiply -36 by -1/36 to move -36 out of the brackets.

$y = - \frac{1}{36} \left({x}^{2} + 12 + 36\right) + 2 \left(- 36\right) \cdot \left(- \frac{1}{36}\right)$

7. Simplify.

$y = - \frac{1}{36} \left({x}^{2} + 12 + 36\right) + 2 \left[\left(- \textcolor{red}{\cancel{\textcolor{b l a c k}{36}}}\right) \cdot \left(- \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{36}}}}\right)\right]$

$y = - \frac{1}{36} \left({x}^{2} + 12 + 36\right) + 2 + 1$

$y = - \frac{1}{36} \left({x}^{2} + 12 + 36\right) + 3$

8. Factor the perfect square trinomial.
The final step to finding the vertex is to factor the perfect square trinomial. This will tell you the $x$ coordinate of the vertex. The $y$ coordinate of the vertex, $3$, has already been found.

$y = - \frac{1}{36} {\left(x + 6\right)}^{2} + 3$

$\therefore$, the vertex is $\left(- 6 , 3\right)$.