What is the axis 0=(y-1)^2-16(x+4) ?

The axis 0=(y-1)^2-16(x+4) is y=1

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axis 0=(y-1)^2-16(x+4)

axis

0 = (y - 1)^{2} - 16 (x + 4)

y = 1

Solution steps

0 = (y - 1)^{2} - 16 (x + 4)

Rewrite

0 = (y - 1)^{2} - 16 (x + 4)

in the standard form:

4 \cdot 4 (x - (- 4)) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (- 4, 1), p = 4

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 1

y = - 1 + 3 x + x^{2}

y = 7 x^{2} - 5 x + 9

y = (x + 2)^{2}

y = - 20 x^{2} + 20000 x - 1800000

(x + 4)^{2} = - 6 (y - 3)