en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f(x)=(x^2-3x+2)/(2x^2+5x-1)

Solution

f (x) = \frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1}

Solution

Domain : x < \frac{- 5 - 33}{4} or \frac{- 5 - 33}{4} < x < \frac{- 5 + 33}{4} or x > \frac{- 5 + 33}{4}

Range : f (x) \leq - \frac{2 34}{11 3} - \frac{7}{11} or f (x) \geq \frac{2 34}{11 3} - \frac{7}{11}

X Intercepts : (2, 0), (1, 0), Y Intercepts : (0, - 2)

Asymptotes : Vertical x = \frac{- 5 - 33}{4}, x = \frac{- 5 + 33}{4}, Horizontal y = \frac{1}{2}

Extreme Points : Maximum (\frac{5 - 102}{11}, - \frac{21 + 2 102}{33}), Minimum (\frac{5 + 102}{11}, - \frac{21 - 2 102}{33})

+1

Interval Notation

Domain : (- \infty, \frac{- 5 - 33}{4}) \cup (\frac{- 5 - 33}{4}, \frac{- 5 + 33}{4}) \cup (\frac{- 5 + 33}{4}, \infty)

Range : (- \infty, - \frac{2 34}{11 3} - \frac{7}{11}] \cup [\frac{2 34}{11 3} - \frac{7}{11}, \infty)

Solution steps

Domain of

\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1} : x < \frac{- 5 - 33}{4} or \frac{- 5 - 33}{4} < x < \frac{- 5 + 33}{4} or x > \frac{- 5 + 33}{4}

Range of

\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1} : f (x) \leq - \frac{2 34}{11 3} - \frac{7}{11} or f (x) \geq \frac{2 34}{11 3} - \frac{7}{11}

Axis interception points of

\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1} :

X Intercepts

: (2, 0), (1, 0),

Y Intercepts

: (0, - 2)

Asymptotes of

\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1} :

Vertical

: x = \frac{- 5 - 33}{4}, x = \frac{- 5 + 33}{4},

Horizontal

: y = \frac{1}{2}

Extreme Points of

\frac{x ^{2} - 3 x + 2}{2 x ^{2} + 5 x - 1} :

Maximum

(\frac{5 - 102}{11}, - \frac{21 + 2 102}{33}),

Minimum

(\frac{5 + 102}{11}, - \frac{21 - 2 102}{33})

Popular Examples

f(x)=4cos^2(x)-3

f (x) = 4 cos^{2} (x) - 3

f(x)=4cos^2(x)-1

f (x) = 4 cos^{2} (x) - 1

P(s)=-10s^2+700s-6000

P (s) = - 10 s^{2} + 700 s - 6000

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

f(x)=(x^2-4x-5)/(x-3)

f (x) = \frac{x ^{2} - 4 x - 5}{x - 3}

Frequently Asked Questions (FAQ)

What is the f(x)=(x^2-3x+2)/(2x^2+5x-1) ?
The f(x)=(x^2-3x+2)/(2x^2+5x-1) is