What is the vertices y=+16-(20x+15)^2 ?

The vertices y=+16-(20x+15)^2 is Maximum (-3/4 ,16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=+16-(20x+15)^2

vertices

y = + 16 - (20 x + 15)^{2}

Maximum (- \frac{3}{4}, 16)

Solution steps

y = 16 - (20 x + 15)^{2}

Rewrite

y = + 16 - (20 x + 15)^{2}

in the form

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

y = - 400 x^{2} - 600 x - 209

The parabola parameters are:

a = - 400, b = - 600, c = - 209

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

x_{v} = - \frac{( - 600 )}{2 ( - 400 )}

Simplify

- \frac{- 600}{2 ( - 400 )} : - \frac{3}{4}

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(- \frac{3}{4}, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 400

Maximum (- \frac{3}{4}, 16)

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

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

3 x^{2} - x + 5

y = x^{2} - 8 x + 17

y = 2 x^{2} - 4 x + 15

What is the vertices y=+16-(20x+15)^2 ?
The vertices y=+16-(20x+15)^2 is Maximum (-3/4 ,16)