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

Ask by Owen Goodwin. in the United States
Mar 19,2025

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The intercepts of \( R(x) \) are at \( x = \frac{3}{2} \) and \( x = 1.5 \). The vertical asymptotes of \( q(x) \) are at \( x = -1 \) and \( x = 0.6 \).

Solución

To find the intercepts of the function \( R(x) = \frac{2x-3}{5x^2+2x-3} \), we need to find the values of \( x \) where the function intersects the x-axis. This means that the function will be equal to zero at these points. To find the vertical asymptotes of the function \( q(x) = \frac{2x-3}{5x^2+2x-3} \), we need to find the values of \( x \) where the denominator of the function is equal to zero. These values will give us the vertical asymptotes of the function. Let's start by finding the intercepts of the function \( R(x) \) and then the vertical asymptotes of the function \( q(x) \). Solve the equation by following steps: - step0: Solve for \(x\): \(2x-3=0\) - step1: Move the constant to the right side: \(2x=0+3\) - step2: Remove 0: \(2x=3\) - step3: Divide both sides: \(\frac{2x}{2}=\frac{3}{2}\) - step4: Divide the numbers: \(x=\frac{3}{2}\) The intercepts of the function \( R(x) = \frac{2x-3}{5x^2+2x-3} \) are \( x = \frac{3}{2} \) and \( x = 1.5 \). Now, let's find the vertical asymptotes of the function \( q(x) = \frac{2x-3}{5x^2+2x-3} \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(5x^{2}+2x-3=0\) - step1: Factor the expression: \(\left(x+1\right)\left(5x-3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&5x-3=0\\&x+1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{3}{5}\\&x=-1\end{align}\) - step4: Rewrite: \(x_{1}=-1,x_{2}=\frac{3}{5}\) The vertical asymptotes of the function \( q(x) = \frac{2x-3}{5x^2+2x-3} \) are \( x = -1 \) and \( x = 0.6 \). Therefore, the intercepts of the function \( R(x) \) are \( x = \frac{3}{2} \) and \( x = 1.5 \), and the vertical asymptotes of the function \( q(x) \) are \( x = -1 \) and \( x = 0.6 \).

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The Deep Dive

To find the intercepts of the function \( R(x) = \frac{2x - 3}{5x^{2} + 2x - 3} \), you should set the numerator equal to zero. This gives you the \( x \)-intercept \( x = \frac{3}{2} \). To find the \( y \)-intercept, substitute \( x = 0 \) into the function, resulting in \( R(0) = -\frac{3}{3} = -1 \). Thus, the intercepts are at \( \left(\frac{3}{2}, 0\right) \) and \( (0, -1) \). To find the vertical asymptotes of \( q(x) = \frac{2x - 3}{5x^{2} + 2x - 3} \), you need to determine when the denominator is zero. Setting \( 5x^{2} + 2x - 3 = 0 \) and solving this quadratic using the quadratic formula gives \( x = \frac{-2 - \sqrt{64}}{10} \) and \( x = \frac{-2 + \sqrt{64}}{10} \). This results in vertical asymptotes at \( x = -1 \) and \( x = \frac{3}{5} \). Want to know something fascinating? Understanding intercepts and asymptotes is crucial not only for mathematics but also for fields like engineering. For example, structural engineers need to calculate loads at specific points, which can be visualized similarly to finding intercepts on a curve. Imagine designing a bridge where every load point needs precision, just like plotting those intercepts! If you’re curious about more sophisticated techniques in calculus, check out L'Hôpital's Rule, which can elegantly handle limits involving indeterminate forms, especially useful when approaching vertical asymptotes. Dive deeper into calculus textbooks or online resources, and you’ll discover techniques that open doors to understanding complex functions and their behaviors. Happy calculating!

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