What is the axis (x-1)^2=8(y+3) ?

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axis (x-1)^2=8(y+3)

axis

(x - 1)^{2} = 8 (y + 3)

x = 1

Solution steps

(x - 1)^{2} = 8 (y + 3)

Rewrite

(x - 1)^{2} = 8 (y + 3)

in the standard form:

4 \cdot 2 (y - (- 3)) = (x - 1)^{2}

Therefore parabola properties are:

(h, k) = (1, - 3), p = 2

Parabola is of the form

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

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = 1

y = 5 x^{2} + 4 x - 3

y = - x^{2} + 10

y = - x^{2} + 12

(x - 3)^{2} = 20 (y + \frac{119}{20})

x > - (y + 4)^{2}