Question 3
Find all vertical asymptote(s) of the function

Ask by Paul Wang. in the United States

Mar 19,2025

Question

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

UpStudy AI · Accepted Answer

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:
$$
x = \frac{3}{5} \quad \text{and} \quad x = -1.
$$
The vertical asymptotes are at $$ x = \frac{3}{5} $$ and $$ x = -1 $$.

Question 3
Find all vertical asymptote(s) of the function

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Calculus Questions