Find the axis of symmetry for this parabola y=-4x^2-8x-6

To find the axis of symmetry, you need to know the vertex cuz the x value of the vertex corresponds to the equation for the axis of symmetry.  Find the vertex by completing the square and putting it into vertex form. Start by setting the whole thing equal to 0. -4x^2 - 8x - 6 = 0.  Next move the 6 over to the other side (which actually breaks several mathematical rules all at the same time! But that's the way to do this, so let's be rebels). Next, the only way we can complete the square is by having the coefficient on the x^2 term be a 1, which currently it is not.  It is a -4. So factor out a -4, and what you have now is: -4(x^2 + 2x) = 6. Now take half of the linear term (which is the 2 on the 2x) and square it.  Half of 2 is 1, and squaring 1 gives us a 1. So that's what we add into the parenthesis on the left: -4(x^2 + 2x + 1) = 6.  BUT because this is an equation, if we do one thing to one side, we also have to do it to the other side.  BUT with that -4 stuck out front, we didn't just add in a 1, we actually added in (-4 * 1) which is a -4. So that's what we even out this equation with on the right: -4(x^2 + 2x + 1) = 6 - 4. Now do the math on the right to get -4(x^2 + 2x + 1) = 2.  The reason we completed the square was to create a perfect binomial on the left: -4(x + 1)^2 = 2. Let's move the 2 over to the left to make this official: -4(x + 1)^2 -2 = y is the equation for that parabola, and the vertex is (-1, -2). Our axis of symmetry is the line that goes through that x value: x = -1

The answer would be d i think but i’m for sure it is not a and c