What is the vertices x=-5y^2 ?

The vertices x=-5y^2 is (h,k)=(0,0),p=-1/20

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x=-5y^2

vertices

x = - 5 y^{2}

(h, k) = (0, 0), p = - \frac{1}{20}

Solution steps

x = - 5 y^{2}

Rewrite

x = - 5 y^{2}

in the standard form:

4 (- \frac{1}{20}) (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{1}{20}

y = 6 x^{2} - 15 x

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

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

x^{2} - 4 x

t^{2} + 2 t - 48