What is the axis y^2-8x-16=0 ?

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axis y^2-8x-16=0

axis

y^{2} - 8 x - 16 = 0

y = 0

Solution steps

y^{2} - 8 x - 16 = 0

Rewrite

y^{2} - 8 x - 16 = 0

in the standard form:

4 \cdot 2 (x - (- 2)) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (- 2, 0), p = 2

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 = 0

(x + 7)^{2} = 5 (y + 3)

x \leq + y^{2}

y = - x^{2} + 80 x - 700

y = \frac{1}{5} (x - 3)^{2} - 5

y = x^{2} - 2 x + 4