Find the equation of the line of symmetry.

Equation: y=a(x-6)(x+12)

it's just the vertical line halfway between the roots, at

x = (6-12)/2 = -3

To find the equation of the line of symmetry for the given equation, we can start by using the standard form of a quadratic equation which is y = ax^2 + bx + c.

In the given equation, y = a(x-6)(x+12), we can simplify it to expand the right side:

y = a(x^2 + 12x - 6x - 72)
= a(x^2 + 6x - 72)

Now let's compare this equation with the standard form:

y = ax^2 + bx + c

We can see that b = 6, indicating that the coefficient of the linear term is 6. Since the linear term is the term without x^2, we can find the x-coordinate of the vertex, which is the line of symmetry, using the formula: x = -b / (2a).

In our case, a = 1 (the coefficient of x^2), and b = 6. Plugging these values into the formula, we get:

x = -6 / (2 * 1)
= -6 / 2
= -3

Therefore, the x-coordinate of the line of symmetry is -3. To find the equation of the line of symmetry, we simply use the formula: x = a, where a is the x-coordinate.

Hence, the equation of the line of symmetry is x = -3.