What is the y= 1/(4x^2-3x+9) ?

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y= 1/(4x^2-3x+9)

y = \frac{1}{4 x ^{2} - 3 x + 9}

Domain : - \infty < x < \infty

Range : 0 < f (x) \leq \frac{16}{135}

Y Intercepts : (0, \frac{1}{9})

Asymptotes : Horizontal y = 0

Extreme Points : Maximum (\frac{3}{8}, \frac{16}{135})

Interval Notation

Domain : (- \infty, \infty)

Solution steps

Domain of

\frac{1}{4 x ^{2} - 3 x + 9} : - \infty < x < \infty

Range of

\frac{1}{4 x ^{2} - 3 x + 9} : 0 < f (x) \leq \frac{16}{135}

Axis interception points of

\frac{1}{4 x ^{2} - 3 x + 9} :

Y Intercepts

: (0, \frac{1}{9})

Asymptotes of

\frac{1}{4 x ^{2} - 3 x + 9} :

Horizontal

: y = 0

Extreme Points of

\frac{1}{4 x ^{2} - 3 x + 9} :

Maximum

(\frac{3}{8}, \frac{16}{135})

f (x) = x^{3} + \frac{3}{2} x^{2} - 6 x

\frac{x ^{10} + x ^{10} + x ^{10}}{x ^{10}}, x \neq = 0

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

y = \frac{x - 1}{x ^{2} - 3}

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