What is the vertices f(x)=2x^2+3x-5 ?

The vertices f(x)=2x^2+3x-5 is Minimum (-3/4 ,-49/8)

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-5

vertices

f (x) = 2 x^{2} + 3 x - 5

Minimum (- \frac{3}{4}, - \frac{49}{8})

Solution steps

y = 2 x^{2} + 3 x - 5

The parabola parameters are:

a = 2, b = 3, c = - 5

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{49}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, - \frac{49}{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{49}{8})

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

4 x - y^{2} + y - \frac{31}{4} = 0

f (x) = x^{2} - 10 x + 16

- 3 x^{2} - 5

f (x) = x^{2} - 8 x - 9

What is the vertices f(x)=2x^2+3x-5 ?
The vertices f(x)=2x^2+3x-5 is Minimum (-3/4 ,-49/8)