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Identify the asymptotes. \[ v(x)=\frac{5 x-6}{5 x^{2}+12 x-9} \] Part: \( 0 / 3 \) Part 1 of 3 Equation(s) of vertical asymptote(s):

Ask by Vega French. in the United States
Mar 18,2025

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The vertical asymptotes are at \( x = \frac{3}{5} \) and \( x = -3 \).

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To find the vertical asymptotes of the function \( v(x) = \frac{5x - 6}{5x^2 + 12x - 9} \), you need to set the denominator equal to zero and solve for \( x \). First, factor the quadratic in the denominator: \[ 5x^2 + 12x - 9 = 0 \] To solve this, we can either factor it directly or apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In this case, applying the quadratic formula with \( a = 5, b = 12, c = -9 \): 1. Calculate the discriminant: \[ b^2 - 4ac = 12^2 - 4 \cdot 5 \cdot (-9) = 144 + 180 = 324 \] 2. Find the roots: \[ x = \frac{-12 \pm \sqrt{324}}{2 \cdot 5} = \frac{-12 \pm 18}{10} \] Solving this gives: - \( x = \frac{6}{10} = \frac{3}{5} \) - \( x = \frac{-30}{10} = -3 \) Thus, the equations of the vertical asymptotes are: **Vertical asymptotes at:** \( x = \frac{3}{5}, x = -3 \)

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