What is the vertices f(x)=-5/8 x^2+1/4 x-2 ?

The vertices f(x)=-5/8 x^2+1/4 x-2 is Maximum (1/5 ,-79/40)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=-5/8 x^2+1/4 x-2

vertices

f (x) = - \frac{5}{8} x^{2} + \frac{1}{4} x - 2

Maximum (\frac{1}{5}, - \frac{79}{40})

Solution steps

y = - \frac{5}{8} x^{2} + \frac{1}{4} x - 2

Rewrite

y = - \frac{5}{8} x^{2} + \frac{1}{4} x - 2

in the form

y = a x^{2} + b x + c

y = - \frac{5 x ^{2}}{8} + \frac{x}{4} - 2

The parabola parameters are:

a = - \frac{5}{8}, b = \frac{1}{4}, c = - 2

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

x_{v} = - \frac{( \frac{1}{4} )}{2 ( - \frac{5}{8} )}

Simplify

- \frac{\frac{1}{4}}{2 ( - \frac{5}{8} )} : \frac{1}{5}

x_{v} = \frac{1}{5}

Plug in

x_{v} = \frac{1}{5}

to find the

y_{v}

value

y_{v} = - \frac{79}{40}

Therefore the parabola vertex is

(\frac{1}{5}, - \frac{79}{40})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{5}{8}

Maximum (\frac{1}{5}, - \frac{79}{40})

f (x) = - 16 x^{2} + 4 x + 10

y = (x + 1)^{2}

- x^{2} + 4 x + 5

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

y = 2.6 + 2.2 x - 0.6 x^{2}

What is the vertices f(x)=-5/8 x^2+1/4 x-2 ?
The vertices f(x)=-5/8 x^2+1/4 x-2 is Maximum (1/5 ,-79/40)