What is the vertices y^2=16x-8 ?

The vertices y^2=16x-8 is (h,k)=(1/2 ,0),p=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y^2=16x-8

vertices

y^{2} = 16 x - 8

(h, k) = (\frac{1}{2}, 0), p = 4

Solution steps

y^{2} = 16 x - 8

Rewrite

y^{2} = 16 x - 8

in the standard form:

4 \cdot 4 (x - \frac{1}{2}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{2}, 0), p = 4

y = - x^{2} + 3

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

y = 5 x^{2} + 13 x - 6

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

x^{2} - 12 x + 44