What is the axis y^2-6y-4x+5=0 ?

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axis y^2-6y-4x+5=0

axis

y^{2} - 6 y - 4 x + 5 = 0

y = 3

Solution steps

y^{2} - 6 y - 4 x + 5 = 0

Rewrite

y^{2} - 6 y - 4 x + 5 = 0

in the standard form

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

Therefore parabola properties are:

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

Parabola is of the form

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

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 3

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

y = 7 x^{2} + 14 x

f (x) = x^{2} + 6 x + 8

y = 2 x^{2} + 12 x + 10

y = 290 (8 x + 0.5 x^{2} - 1.4)