What is the vertices f(x)=-x^2+4x+2 ?

The vertices f(x)=-x^2+4x+2 is Maximum (2,6)

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+4x+2

vertices

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

Maximum (2, 6)

Solution steps

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

The parabola parameters are:

a = - 1, b = 4, c = 2

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

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(2, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 6)

f (x) = x^{2} + 8 + 15

y = x^{2} + 4 x - 5

3 x^{2} + 1

y = 7 x^{2} + 5

- 2 x^{2} - 4 x + 3

What is the vertices f(x)=-x^2+4x+2 ?
The vertices f(x)=-x^2+4x+2 is Maximum (2,6)