Graph the fational function- \[ f(x)=\frac{-10 x+20}{x^{2}-3 x-10} \] Start by drawing the vertical and horizontal asympiotes. Then plot the intercepts (il ony), and plot at least one point on each side of each vertical asymptote. Finally, click on the graph-a-function button.

Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=-10x+\frac{20}{x^{2}-3x-10}\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=-10\times 0+\frac{20}{0^{2}-3\times 0-10}\) - step2: Multiply: \(f\left(0\right)=-10\times 0+\frac{20}{0^{2}-0-10}\) - step3: Multiply: \(f\left(0\right)=0+\frac{20}{0^{2}-0-10}\) - step4: Simplify: \(f\left(0\right)=-2\) Find the horizontal asymptotes of \( f(x)=-10x+20/(x^2-3x-10) \). Function by following steps: - step0: Find the horizontal asymptotes: \(f\left(x\right)=-10x+\frac{20}{x^{2}-3x-10}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(f\left(x\right)\right)\) and \(\lim _{x\rightarrow -\infty}\left(f\left(x\right)\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(-10x+\frac{20}{x^{2}-3x-10}\right)\\&\lim _{x\rightarrow -\infty}\left(-10x+\frac{20}{x^{2}-3x-10}\right)\end{align}\) - step2: Calculate: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step3: The function has no horizontal asymptotes: \(\textrm{No horizontal asymptotes}\) Analyze the x intercept of the function \( f(x)=-10x+20/(x^2-3x-10) \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=-10x+\frac{20}{x^{2}-3x-10}\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=-10x+\frac{20}{x^{2}-3x-10}\) - step2: Swap the sides: \(-10x+\frac{20}{x^{2}-3x-10}=0\) - step3: Find the domain: \(-10x+\frac{20}{x^{2}-3x-10}=0,x \in \left(-\infty,-2\right)\cup \left(-2,5\right)\cup \left(5,+\infty\right)\) - step4: Multiply both sides of the equation by LCD: \(\left(-10x+\frac{20}{x^{2}-3x-10}\right)\left(x^{2}-3x-10\right)=0\times \left(x^{2}-3x-10\right)\) - step5: Simplify the equation: \(-10x^{3}+30x^{2}+100x+20=0\) - step6: Factor the expression: \(-10\left(x^{3}-3x^{2}-10x-2\right)=0\) - step7: Divide both sides: \(x^{3}-3x^{2}-10x-2=0\) - step8: Calculate: \(\begin{align}&x\approx 5.05606\\&x\approx -1.841222\\&x\approx -0.214838\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x\approx 5.05606\\&x\approx -1.841222\\&x\approx -0.214838\end{align},x \in \left(-\infty,-2\right)\cup \left(-2,5\right)\cup \left(5,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&x\approx 5.05606\\&x\approx -1.841222\\&x\approx -0.214838\end{align}\) - step11: Calculate: \(x_{1}\approx -1.841222,x_{2}\approx -0.214838,x_{3}\approx 5.05606\) Find the vertical asymptotes of \( f(x)=-10x+20/(x^2-3x-10) \). Function by following steps: - step0: Find the vertical asymptotes: \(f\left(x\right)=-10x+\frac{20}{x^{2}-3x-10}\) - step1: Evaluate the limit: \(\begin{align}&\lim _{x\rightarrow -2}\left(-10x+\frac{20}{x^{2}-3x-10}\right)\\&\lim _{x\rightarrow 5}\left(-10x+\frac{20}{x^{2}-3x-10}\right)\end{align}\) - step2: Calculate: \(\begin{align}&\textrm{The limit does not exist}\\&\textrm{The limit does not exist}\end{align}\) - step3: \(x=-2\) is a vertical asymptote\(:\) \(\begin{align}&x=-2\textrm{ }\textrm{is a vertical asymptote}\\&x=5\textrm{ }\textrm{is a vertical asymptote}\end{align}\) - step4: List all vertical asymptotes of the function: \(\begin{align}&x=-2\\&x=5\end{align}\) Let's analyze the function \( f(x) = \frac{-10x + 20}{x^2 - 3x - 10} \) step by step. ### Step 1: Find Vertical Asymptotes The vertical asymptotes occur where the denominator is zero. From our calculations, we found: - The vertical asymptotes are at \( x = -2 \) and \( x = 5 \). ### Step 2: Find Horizontal Asymptotes Horizontal asymptotes are determined by the degrees of the numerator and denominator. In this case: - The degree of the numerator (linear) is 1. - The degree of the denominator (quadratic) is 2. Since the degree of the denominator is greater than the degree of the numerator, there are **no horizontal asymptotes**. ### Step 3: Find Intercepts #### X-Intercepts To find the x-intercepts, we set the numerator equal to zero: \[ -10x + 20 = 0 \implies x = 2 \] However, we also found additional x-intercepts through calculations: - \( x_1 \approx -1.841222 \) - \( x_2 \approx -0.214838 \) - \( x_3 \approx 5.05606 \) #### Y-Intercept To find the y-intercept, we evaluate \( f(0) \): \[ f(0) = \frac{-10(0) + 20}{0^2 - 3(0) - 10} = \frac{20}{-10} = -2 \] Thus, the y-intercept is at \( (0, -2) \). ### Step 4: Plot Points Around Vertical Asymptotes We need to plot at least one point on each side of the vertical asymptotes \( x = -2 \) and \( x = 5 \). 1. **For \( x = -2 \)**: - Choose \( x = -3 \): \[ f(-3) = \frac{-10(-3) + 20}{(-3)^2 - 3(-3) - 10} = \frac{30 + 20}{9 + 9 - 10} = \frac{50}{8} = 6.25 \] - Choose \( x = -1 \): \[ f(-1) = \frac{-10(-1) + 20}{(-1)^2 - 3(-1) - 10} = \frac{10 + 20}{1 + 3 - 10} = \frac{30}{-6} = -5 \] 2. **For \( x = 5 \)**: - Choose \( x = 4 \): \[ f(4) = \frac{-10(4) + 20}{(4)^2 - 3(4) - 10} = \frac{-40 + 20}{16 - 12 - 10} = \frac{-20}{-6} \approx 3.33 \] - Choose \( x = 6 \): \[ f(6) = \frac{-10(6) + 20}{(6)^2 - 3(6) - 10} = \frac{-60 + 20}{36 - 18 - 10} = \frac{-40}{8} = -5 \] ### Summary of Points to Plot - Vertical Asymptotes: \( x = -2 \), \( x = 5 \) - X-Intercepts: \( x \approx -1.841222, -0.214838, 2, 5.05606 \) - Y-Intercept: \( (0, -2) \) - Points: - Near \( x = -2 \): \( (-3, 6.25) \) and \( (-1, -5) \) - Near \( x = 5 \): \( (4, 3.33) \) and \( (6, -5) \) Now, let's graph the function with the identified asymptotes, intercepts, and additional points.