What is the vertices y=16x^2-24x+9 ?

The vertices y=16x^2-24x+9 is Minimum (3/4 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=16x^2-24x+9

vertices

y = 16 x^{2} - 24 x + 9

Minimum (\frac{3}{4}, 0)

Solution steps

y = 16 x^{2} - 24 x + 9

The parabola parameters are:

a = 16, b = - 24, c = 9

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

x_{v} = - \frac{( - 24 )}{2 \cdot 16}

Simplify

- \frac{- 24}{2 \cdot 16} : \frac{3}{4}

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

Plug in

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

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(\frac{3}{4}, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 16

Minimum (\frac{3}{4}, 0)

(- 4, 0), (5, 9)

x^{2} - 8 x + 15

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

(- 5.9) (8.14)

(4, - 1), (- 1, 1)

What is the vertices y=16x^2-24x+9 ?
The vertices y=16x^2-24x+9 is Minimum (3/4 ,0)