What is the asymptotes of f(x)=((16x^2-8x+1))/(4x-3) ?

The asymptotes of f(x)=((16x^2-8x+1))/(4x-3) is Vertical: x= 3/4 ,Slant: y=4x+1

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asymptotes of f(x)=((16x^2-8x+1))/(4x-3)

asymptotes

f (x) = \frac{( 16 x ^{2} - 8 x + 1 )}{4 x - 3}

Vertical : x = \frac{3}{4}, Slant : y = 4 x + 1

Solution steps

Vertical asymptotes of

\frac{16 x ^{2} - 8 x + 1}{4 x - 3} : x = \frac{3}{4}

Horizontal Asymptotes of

\frac{16 x ^{2} - 8 x + 1}{4 x - 3} :

None

Slant Asymptotes of

\frac{16 x ^{2} - 8 x + 1}{4 x - 3} : y = 4 x + 1

Vertical : x = \frac{3}{4}, Slant : y = 4 x + 1

f (x) = 16 - t

f (x) = \frac{e ^{x^{8}}}{2}

f (x) = \frac{( 3 x + 1 )}{( 4 x + 2 )}

\frac{x + 1}{( x + 2 ) ^{2}}

x - 1 + 3

What is the asymptotes of f(x)=((16x^2-8x+1))/(4x-3) ?
The asymptotes of f(x)=((16x^2-8x+1))/(4x-3) is Vertical: x= 3/4 ,Slant: y=4x+1