en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f(t)=t^2+t

Solution

f (t) = t^{2} + t

Solution

Domain : - \infty < t < \infty

Range : f (t) \geq - \frac{1}{4}

X Intercepts : (0, 0), (- 1, 0), Y Intercepts : (0, 0)

Vertex : Minimum (- \frac{1}{2}, - \frac{1}{4})

Inverse : \frac{- 1 + 1 + 4 t}{2}, \frac{- 1 - 1 + 4 t}{2}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : [- \frac{1}{4}, \infty)

Solution steps

Domain of

t^{2} + t : - \infty < t < \infty

Range of

t^{2} + t : f (t) \geq - \frac{1}{4}

Axis interception points of

t^{2} + t :

X Intercepts

: (0, 0), (- 1, 0),

Y Intercepts

: (0, 0)

Vertex of

t^{2} + t :

Minimum

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

Inverse of

t^{2} + t : \frac{- 1 + 1 + 4 t}{2}, \frac{- 1 - 1 + 4 t}{2}

Popular Examples

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

f(x)=2x^2+12x+7

f (x) = 2 x^{2} + 12 x + 7

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

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

y=x^3-2x^2+16x-32

y = x^{3} - 2 x^{2} + 16 x - 32

p(t)=(50)/(1+80e^{-t)}

p (t) = \frac{50}{1 + 80 e ^{- t}}

Frequently Asked Questions (FAQ)

What is the f(t)=t^2+t ?
The f(t)=t^2+t is