What is the axis x^2-2x-4y+17=0 ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

axis x^2-2x-4y+17=0

axis

x^{2} - 2 x - 4 y + 17 = 0

x = 1

Solution steps

x^{2} - 2 x - 4 y + 17 = 0

Rewrite

x^{2} - 2 x - 4 y + 17 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (1, 4), p = 1

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 = 1

y = 2 x^{2} - 3 x

y = 3 x^{2} - 6 x - 24

y = 4 x^{2} + 8 x + 13

16 x^{2} - 72 x - 112 y = - 221

x = (4 y + 3) (2 y - 5)