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

The axis f(x)=-x^2-4x+5 is x=-2

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axis f(x)=-x^2-4x+5

axis

f (x) = - 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 (- \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

x = y^{2} - 2 y

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

y = x^{2} - 2 x^{1}

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

y = 12 x^{2} - 72 x + 77