What is the vertices 2y^2+8y+x+1=0 ?

The vertices 2y^2+8y+x+1=0 is (h,k)=(7,-2),p=-1/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2y^2+8y+x+1=0

vertices

2 y^{2} + 8 y + x + 1 = 0

(h, k) = (7, - 2), p = - \frac{1}{8}

Solution steps

2 y^{2} + 8 y + x + 1 = 0

Rewrite

2 y^{2} + 8 y + x + 1 = 0

in the standard form

4 (- \frac{1}{8}) (x - 7) = (y - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (7, - 2), p = - \frac{1}{8}

y^{2} - 12 y - 4 x + 28 = 0

f (x) = (x - 2)^{2} - 9

(3 x - 4) (4 x - 3) - 2 x

4 x^{2}

y = x^{2} - 2

What is the vertices 2y^2+8y+x+1=0 ?
The vertices 2y^2+8y+x+1=0 is (h,k)=(7,-2),p=-1/8