what is the axis of symmetry of y=-4(x+8)^2-6
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what is the axis of symmetry of y=-4(x+8)^2-6.

The right hand side here of this equation is a quadratic and all quadratics are U-shaped.

The coefficient of the x^2-term is -4, a negative value, hence the quadratic is a parabola that is U-shaped downwards, like an umbrella.

The axis of symmetry is a vertical line that passes through the vertex of the parabola, or U-shape.

There are two ways of finding the axis of symmetry. One way is by differentiation to find the maximum (highest) point on the graph (the turning point), which will give the location of the vertex.

y = -4(x+8)^2 - 6

dy/dx = -8(x+8)

The slope, dy/dx, is zero at x = -8.

The turning point, which here is a maximum, occurs at x = -8, so the axis of symmetry is the vertical line, x = -8.

Answer: axis of symmetry is x = -8

 

The other way is by completeing the square, but that is already done in our case, so half our work is already done.

We have  y=-4(x+8)^2-6, or

(y + 6) = -4(x+8)^2, i.e.

Y = AX^2,  where Y = y+6, A = -4, X = x+8.

The origin for the X-Y coordinates is X=0, Y=0 and this is the vertex of the parabola and is where the line of symmetry passes through.

But X = 0 means x+8 = 0. i.e. x = -8.

And Y = 0 means y+6 = 0. i.e. y = -6.

So the vertex is (-8, -6) and the axis of symmetry is the vertical line passing through the vertex which is x = -8.

Answer: axis of symmetry is x = -8

by Level 11 User (81.5k points)
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