What is the axis y=(x-5)^2-4 ?

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axis y=(x-5)^2-4

axis

y = (x - 5)^{2} - 4

x = 5

Solution steps

y = (x - 5)^{2} - 4

Rewrite

y = (x - 5)^{2} - 4

in the standard form:

4 \cdot \frac{1}{4} (y - (- 4)) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, - 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 = 5

y = - 3 x^{2} - 2

y = - \frac{1}{11} (x + 5)^{2} + 5

y = - 2 x + x (x - 2) - 2 x (x - 1) - (x - 3)

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

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