Question 3 Find all vertical asymptote(s) of the function \( q(x)=\frac{2 x-3}{5 x^{2}+2 x-3} \)

We first find the values of \( x \) that make the denominator zero, since these are potential vertical asymptotes. 1. Set the denominator equal to zero: \[ 5x^2 + 2x - 3 = 0 \] 2. Solve this quadratic using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 5 \), \( b = 2 \), and \( c = -3 \). 3. Substitute the values: \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(5)(-3)}}{2(5)} = \frac{-2 \pm \sqrt{4 + 60}}{10} = \frac{-2 \pm \sqrt{64}}{10} \] 4. Simplify the square root: \[ x = \frac{-2 \pm 8}{10} \] 5. Find the two solutions: For the positive branch: \[ x = \frac{-2 + 8}{10} = \frac{6}{10} = \frac{3}{5} \] For the negative branch: \[ x = \frac{-2 - 8}{10} = \frac{-10}{10} = -1 \] 6. Verify that these values do not cancel with the numerator \( 2x - 3 \). Since \[ 2\left(\frac{3}{5}\right) - 3 = \frac{6}{5} - 3 \neq 0 \quad \text{and} \quad 2(-1) - 3 \neq 0, \] neither factor cancels. Thus, the vertical asymptotes are at: \[ x = \frac{3}{5} \quad \text{and} \quad x = -1. \]