What is the vertices y^2+4x-2y+21=0 ?

The vertices y^2+4x-2y+21=0 is (h,k)=(-5,1),p=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y^2+4x-2y+21=0

vertices

y^{2} + 4 x - 2 y + 21 = 0

(h, k) = (- 5, 1), p = - 1

Solution steps

y^{2} + 4 x - 2 y + 21 = 0

Rewrite

y^{2} + 4 x - 2 y + 21 = 0

in the standard form

4 (- 1) (x - (- 5)) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (- 5, 1), p = - 1

f (x) = 415 x^{2} - 4600 x + 200000

y = x^{2} - 4 x 3

f (x) = x^{2} - 3 x - 10

- x^{2} + 2 x - 10

f (x) = 3 x^{2} - 12 x + 13

What is the vertices y^2+4x-2y+21=0 ?
The vertices y^2+4x-2y+21=0 is (h,k)=(-5,1),p=-1