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Vertex and symmetry of y=-x^2-8x-9

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\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{-9} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ -8}{2(-1)}~~~~ ,~~~~ -9-\cfrac{ (-8)^2}{4(-1)}\right) \implies\left( - \cfrac{ -8 }{ -2 }~~,~~-9 - \cfrac{ 64 }{ -4 } \right) \\\\\\ \left( -\cfrac{ 8 }{ 2 }~~,~~-9 + 12 \right)\implies (-4~~,~~3)

this is a parabola whose independent variable is "x", or the squared variable is "x", meaning is a vertical parabola, so its axis of symmetry will come from the vertex's x-coordinate, x = -4.

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