What is 2x^2-5x+k(x)=0 ?

The solution to 2x^2-5x+k(x)=0 is Parabola with (h,k)=(5/4 , 25/8),p=-1/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2-5x+k(x)=0

2 x^{2} - 5 x + k (x) = 0

(h, k) = (\frac{5}{4}, \frac{25}{8}), p = - \frac{1}{8}

Solution steps

2 x^{2} - 5 x + y = 0

Rewrite

2 x^{2} - 5 x + y = 0

in the standard form

4 (- \frac{1}{8}) (y - \frac{25}{8}) = (x - \frac{5}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{4}, \frac{25}{8}), p = - \frac{1}{8}

2 x + y = x^{2}

y^{2} + 2 y - 4 x = 7

x^{2} = - 7 y

y^{2} - 3 y + 5 x + 2 = 0

(x - 3)^{2} = - 4 (y - 5)

What is 2x^2-5x+k(x)=0 ?
The solution to 2x^2-5x+k(x)=0 is Parabola with (h,k)=(5/4 , 25/8),p=-1/8