What is the vertices f(x)=(x-7)^2-1 ?

The vertices f(x)=(x-7)^2-1 is Minimum (7,-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-7)^2-1

vertices

f (x) = (x - 7)^{2} - 1

Minimum (7, - 1)

Solution steps

y = (x - 7)^{2} - 1

Rewrite

y = (x - 7)^{2} - 1

in the form

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

y = x^{2} - 14 x + 48

The parabola parameters are:

a = 1, b = - 14, c = 48

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

x_{v} = - \frac{( - 14 )}{2 \cdot 1}

Simplify

x_{v} = 7

Plug in

x_{v} = 7

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(7, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (7, - 1)

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

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

x^{2} + 2 x

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

5 x^{2} + 20 x - 17

What is the vertices f(x)=(x-7)^2-1 ?
The vertices f(x)=(x-7)^2-1 is Minimum (7,-1)