What is the vertices f(x)=2x^2+9x+10 ?

The vertices f(x)=2x^2+9x+10 is Minimum (-9/4 ,-1/8)

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+9x+10

vertices

f (x) = 2 x^{2} + 9 x + 10

Minimum (- \frac{9}{4}, - \frac{1}{8})

Solution steps

y = 2 x^{2} + 9 x + 10

The parabola parameters are:

a = 2, b = 9, c = 10

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

x_{v} = - \frac{9}{2 \cdot 2}

Simplify

x_{v} = - \frac{9}{4}

Plug in

x_{v} = - \frac{9}{4}

to find the

y_{v}

value

y_{v} = - \frac{1}{8}

Therefore the parabola vertex is

(- \frac{9}{4}, - \frac{1}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{9}{4}, - \frac{1}{8})

f (x) = x^{2} + 6 x - 16

x = 1 - y^{2}

f (x) = 8 x - 13 - x^{2}

- \frac{( x + 3 ) ^{2}}{20} + 1

x^{2} - 2 x + 2

What is the vertices f(x)=2x^2+9x+10 ?
The vertices f(x)=2x^2+9x+10 is Minimum (-9/4 ,-1/8)