What is the vertices x^2+6x+8y=7 ?

The vertices x^2+6x+8y=7 is Maximum (-3,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2+6x+8y=7

vertices

x^{2} + 6 x + 8 y = 7

Maximum (- 3, 2)

Solution steps

x^{2} + 6 x + 8 y = 7

Rewrite

x^{2} + 6 x + 8 y = 7

in the form

y = a x^{2} + b x + c

y = - \frac{x ^{2}}{8} - \frac{3 x}{4} + \frac{7}{8}

The parabola parameters are:

a = - \frac{1}{8}, b = - \frac{3}{4}, c = \frac{7}{8}

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - \frac{3}{4} )}{2 ( - \frac{1}{8} )}

Simplify

- \frac{- \frac{3}{4}}{2 ( - \frac{1}{8} )} : - 3

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 3, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{8}

Maximum (- 3, 2)

x^{2} - 6 x + 5 y = - 34

x^{2} + 2 x

f (x) = 2 (x - 1)^{2} + 4

5 x^{2} + 20 x - 17

y = 2 x^{2} + 24 x - 6