What is the axis y=(x-3)^2-4 ?

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axis y=(x-3)^2-4

axis

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

x = 3

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (3, - 4), p = \frac{1}{4}

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 = 3

(x + 6)^{2} = 12 (y + 1)

y = x^{2} - 6 x - 3

y = 3 x^{2} - 30 x + 77

y = x^{2} - 7

y^{2} = 32 x