What is the vertices f(x)=x^2-2x-15 ?

The vertices f(x)=x^2-2x-15 is Minimum (1,-16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2-2x-15

vertices

f (x) = x^{2} - 2 x - 15

Minimum (1, - 16)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 15

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(1, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 16)

x^{2} - 9 y^{2} = 9

\frac{x ^{2}}{36} + \frac{y ^{2}}{100} = 1

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

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

9 y^{2} - 16 x^{2} - 256 = 0

What is the vertices f(x)=x^2-2x-15 ?
The vertices f(x)=x^2-2x-15 is Minimum (1,-16)