What is the vertices f(x)=x^2+16x+63 ?

The vertices f(x)=x^2+16x+63 is Minimum (-8,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2+16x+63

vertices

f (x) = x^{2} + 16 x + 63

Minimum (- 8, - 1)

Solution steps

y = x^{2} + 16 x + 63

The parabola parameters are:

a = 1, b = 16, c = 63

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

x_{v} = - \frac{16}{2 \cdot 1}

Simplify

x_{v} = - 8

Plug in

x_{v} = - 8

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 8, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 8, - 1)

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

y = (x + 3) (x - 3)

f (x) = - 3 x^{2} - 18 x - 25

y = 18 x^{2} - 14 x + 9

f (x) = - 3 x^{2} + 18 x - 8

What is the vertices f(x)=x^2+16x+63 ?
The vertices f(x)=x^2+16x+63 is Minimum (-8,-1)