What is the equation of the axis of symmetry of the graph of y + 3x – 6 = –3(x – 2)2 + 4?

Answer 1

y + 3x – 6 = –3(x – 2)² + 4

y + 3x – 6 = –3(x² - 4x + 4) + 4

y + 3x – 6 = –3x² + 12x - 12 + 4

y = –3x² + 12x - 3x - 12 + 4 + 6

y = –3x² + 9x - 2

y + 2 = -3(x² - 3x)

y + 2 - 27/4 = -3(x² - 3x + 9/4)

y + 8/4 - 27/4 = -3(x - 3/2)²

y - 19/4 = -3(x - 3/2)²

y = -3(x - 3/2)² + 19/4

So this is a parabola opening down, with a vertical stretch factor of 3 and a vertex at (3/2, 19/4) and a line of symmetry of x = 3/2

Answer 2

I assume what we're looking at is a quadratic, given the context of the question I'll start here:

y+3x-6=-3(x-2)²+4

First, we multiply out the square on the right side

y+3x-6=-3(x²-4x+4)+4

Distribute the -3

y+3x-6=-3x²+12x-12+4

Isolate the y and combine like terms

y = -3x²+9x-2

Now we have it in a form where we use the shortcut that says the axis of symmetry is x=-b/(2a)

So the answer is

x = -9/(2(-3)) = 3/2