What is the vertices f(x)=2x^2+3x-1 ?

The vertices f(x)=2x^2+3x-1 is Minimum (-3/4 ,-17/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=2x^2+3x-1

vertices

f (x) = 2 x^{2} + 3 x - 1

Minimum (- \frac{3}{4}, - \frac{17}{8})

Solution steps

y = 2 x^{2} + 3 x - 1

The parabola parameters are:

a = 2, b = 3, c = - 1

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

x_{v} = - \frac{3}{2 \cdot 2}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{17}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, - \frac{17}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{3}{4}, - \frac{17}{8})

- 2 x^{2} + 12 x

f (x) = 4 x^{2} - 7 x - 2

2 y^{2} + 8 y + x + 1 = 0

y^{2} - 12 y - 4 x + 28 = 0

f (x) = (x - 2)^{2} - 9

What is the vertices f(x)=2x^2+3x-1 ?
The vertices f(x)=2x^2+3x-1 is Minimum (-3/4 ,-17/8)