Let $f ( x ) = \frac { 2 x - 1 } { 2 x ^ { 2 } + 5 x - 3 }$ The set of all vertical asymptotes of ƒ is

A) $\left\{ x = \frac { 1 } { 2 } \right\}$
B) $\{ x = - 3 \}$
C) $\left\{ x = \frac { 1 } { 2 } , x = - 3 \right\}$
D) $\{ x = 5 \}$
E) $\{ x = - 3 , x = 5 \}$

Question

Let $f ( x ) = \frac { 2 x - 1 } { 2 x ^ { 2 } + 5 x - 3 }$ The set of all vertical asymptotes of ƒ is

A) $\left\{ x = \frac { 1 } { 2 } \right\}$
B) $\{ x = - 3 \}$
C) $\left\{ x = \frac { 1 } { 2 } , x = - 3 \right\}$
D) $\{ x = 5 \}$
E) $\{ x = - 3 , x = 5 \}$

Quizplus · Accepted Answer

Vertical asymptotes occur where the denominator of a rational function is zero (assuming the numerator is not zero at the same point, in which case it might be a hole instead). To find the vertical asymptotes of $f(x) = \frac{2x - 1}{2x^2 + 5x - 3}$, we set the denominator equal to zero and solve for $x$: $2x^2 + 5x - 3 = 0$. Factoring or using the quadratic formula, we find the roots of the denominator to be $x = -3$ and $x = \frac{1}{2}$. However, the numerator $2x - 1$ also equals zero when $x = \frac{1}{2}$, which means the function is undefined at this point but it does not necessarily indicate a vertical asymptote without further analysis. The correct identification of vertical asymptotes requires checking if these points indeed make the function approach infinity. In this case, the simplification or misinterpretation might have led to an incorrect listing of options. The correct approach is to identify where only the denominator is zero, and the numerator is not zero, which leads to $x = -3$ as the correct answer, acknowledging a mistake in the explanation process.

Let f(x)=2x−12x2+5x−3f ( x ) = \frac { 2 x - 1 } { 2 x ^ { 2 } + 5 x - 3 }f(x)=2x2+5x−32x−1​

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