What is the domain of y=(2x^2-5x+3)/(4x-1) ?

The domain of y=(2x^2-5x+3)/(4x-1) is x 1/4

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domain of y=(2x^2-5x+3)/(4x-1)

domain

y = \frac{2 x ^{2} - 5 x + 3}{4 x - 1}

x < \frac{1}{4} or x > \frac{1}{4}

Interval Notation

(- \infty, \frac{1}{4}) \cup (\frac{1}{4}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{1}{4}

The function domain

x < \frac{1}{4} or x > \frac{1}{4}

\frac{2}{2 + 5 x}

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

x + 6 + 5

f (x) = (x)^{1} (x) = 4 x^{3} + x

f (x) = \frac{20 x}{x + 19}

What is the domain of y=(2x^2-5x+3)/(4x-1) ?
The domain of y=(2x^2-5x+3)/(4x-1) is x< 1/4 \lor x> 1/4