What is the vertices+y^2+12x+4y-32=0 ?

The vertices+y^2+12x+4y-32=0 is (h,k)=(3,-2),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+12x+4y-32=0

vertices

+ y^{2} + 12 x + 4 y - 32 = 0

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

Solution steps

y^{2} + 12 x + 4 y - 32 = 0

Rewrite

y^{2} + 12 x + 4 y - 32 = 0

in the standard form

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

Therefore parabola properties are:

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

f (x) = - 2 x^{2} - 16 x - 30

x = - 5 y^{2}

y = 6 x^{2} - 15 x

f (x) = - (x - 5)^{2} - 4

y = \frac{x ^{2} 5 x 8}{x 3}

What is the vertices+y^2+12x+4y-32=0 ?
The vertices+y^2+12x+4y-32=0 is (h,k)=(3,-2),p=-3