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

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axis y=x^2-4x-5

axis

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

x = 2

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- 9)) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, - 9), 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 = 2

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

y = \frac{x ^{4} - 1}{x ^{2} - 1}

y = 64 x^{2}

y = 2.5 + 2.2 x - 0.6 x^{2}

x^{2} = - 14 (y + 1)