What is the vertices y^2-2y-12x+25=0 ?

The vertices y^2-2y-12x+25=0 is (h,k)=(2,1),p=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y^2-2y-12x+25=0

vertices

y^{2} - 2 y - 12 x + 25 = 0

(h, k) = (2, 1), p = 3

Solution steps

y^{2} - 2 y - 12 x + 25 = 0

Rewrite

y^{2} - 2 y - 12 x + 25 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, 1), p = 3

x^{2} - 4 x - 6

y = 84 x - x^{2}

\frac{x ^{2} x ^{1}}{x}

y = x^{2} - 2 x

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

What is the vertices y^2-2y-12x+25=0 ?
The vertices y^2-2y-12x+25=0 is (h,k)=(2,1),p=3