4. \( f(x)=\frac{-2 x^{2}-2 x+4}{2 x+1} \) a. Vertical Asymptote(s) Equation: b. End Behavior Asymptote Equation: chorizontal) \( x \)-intercept(s): d. \( y \)-intercept: e. Coordinate of Hole: f. Domain: g. Range:

Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=-\frac{2\times 0^{2}+2\times 0-4}{2\times 0+1}\) - step2: Multiply: \(f\left(0\right)=-\frac{2\times 0^{2}+0-4}{2\times 0+1}\) - step3: Multiply: \(f\left(0\right)=-\frac{2\times 0^{2}+0-4}{0+1}\) - step4: Simplify: \(f\left(0\right)=4\) Analyze the x intercept of the function \( f(x)=-\frac{2 x^{2}+2 x-4}{2 x+1} \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=-\frac{2x^{2}+2x-4}{2x+1}\) - step2: Swap the sides: \(-\frac{2x^{2}+2x-4}{2x+1}=0\) - step3: Find the domain: \(-\frac{2x^{2}+2x-4}{2x+1}=0,x\neq -\frac{1}{2}\) - step4: Rewrite the expression: \(\frac{-2x^{2}-2x+4}{2x+1}=0\) - step5: Cross multiply: \(-2x^{2}-2x+4=\left(2x+1\right)\times 0\) - step6: Simplify the equation: \(-2x^{2}-2x+4=0\) - step7: Factor the expression: \(-2\left(x-1\right)\left(x+2\right)=0\) - step8: Divide the terms: \(\left(x-1\right)\left(x+2\right)=0\) - step9: Separate into possible cases: \(\begin{align}&x-1=0\\&x+2=0\end{align}\) - step10: Solve the equation: \(\begin{align}&x=1\\&x=-2\end{align}\) - step11: Check if the solution is in the defined range: \(\begin{align}&x=1\\&x=-2\end{align},x\neq -\frac{1}{2}\) - step12: Find the intersection: \(\begin{align}&x=1\\&x=-2\end{align}\) - step13: Calculate: \(x_{1}=-2,x_{2}=1\) Find the horizontal asymptotes of \( f(x)=-\frac{2 x^{2}+2 x-4}{2 x+1} \). Function by following steps: - step0: Find the horizontal asymptotes: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(f\left(x\right)\right)\) and \(\lim _{x\rightarrow -\infty}\left(f\left(x\right)\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(-\frac{2x^{2}+2x-4}{2x+1}\right)\\&\lim _{x\rightarrow -\infty}\left(-\frac{2x^{2}+2x-4}{2x+1}\right)\end{align}\) - step2: Calculate: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step3: The function has no horizontal asymptotes: \(\textrm{No horizontal asymptotes}\) Find the vertical asymptotes of \( f(x)=-\frac{2 x^{2}+2 x-4}{2 x+1} \). Function by following steps: - step0: Find the vertical asymptotes: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Evaluate the limit: \(\lim _{x\rightarrow -\frac{1}{2}}\left(-\frac{2x^{2}+2x-4}{2x+1}\right)\) - step2: Rewrite the fraction: \(\lim _{x\rightarrow -\frac{1}{2}}\left(\frac{-2x^{2}-2x+4}{2x+1}\right)\) - step3: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow -\frac{1}{2}^{-}}\left(\frac{-2x^{2}-2x+4}{2x+1}\right)\\&\lim _{x\rightarrow -\frac{1}{2}^{+}}\left(\frac{-2x^{2}-2x+4}{2x+1}\right)\end{align}\) - step4: Evaluate the left-hand limit: \(\begin{align}&-\infty\\&+\infty\end{align}\) - step5: Limit does not exist: \(\textrm{The limit does not exist}\) - step6: \(x=-\frac{1}{2}\) is a vertical asymptote\(:\) \(\begin{align}&x=-\frac{1}{2}\textrm{ }\textrm{is a vertical asymptote}\end{align}\) - step7: List all vertical asymptotes of the function: \(\begin{align}&x=-\frac{1}{2}\end{align}\) Analyze the range of the function \( f(x)=-\frac{2 x^{2}+2 x-4}{2 x+1} \) Function by following steps: - step0: Find the range: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Evaluate: \(f\left(x\right) \in \mathbb{R}\) Analyze the range of the function \( f(x)=-\frac{2 x^{2}+2 x-4}{2 x+1} \) Function by following steps: - step0: Find the range: \(f\left(x\right)=-\frac{2x^{2}+2x-4}{2x+1}\) - step1: Evaluate: \(f\left(x\right) \in \mathbb{R}\) Let's analyze the function \( f(x) = \frac{-2x^2 - 2x + 4}{2x + 1} \) step by step to find the required characteristics. ### a. Vertical Asymptote(s) Equation: To find the vertical asymptote, we set the denominator equal to zero: \[ 2x + 1 = 0 \implies x = -\frac{1}{2} \] **Vertical Asymptote:** \( x = -\frac{1}{2} \) ### b. End Behavior Asymptote Equation: For rational functions, if the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. In this case, the degree of the numerator (2) is greater than the degree of the denominator (1). **End Behavior Asymptote:** No horizontal asymptotes. ### c. \( x \)-intercept(s): To find the \( x \)-intercepts, we set the numerator equal to zero: \[ -2x^2 - 2x + 4 = 0 \] Factoring or using the quadratic formula gives us: \[ x_{1} = -2, \quad x_{2} = 1 \] **\( x \)-intercepts:** \( x = -2 \) and \( x = 1 \) ### d. \( y \)-intercept: To find the \( y \)-intercept, we evaluate \( f(0) \): \[ f(0) = \frac{-2(0)^2 - 2(0) + 4}{2(0) + 1} = \frac{4}{1} = 4 \] **\( y \)-intercept:** \( 4 \) ### e. Coordinate of Hole: A hole occurs when a factor cancels in the numerator and denominator. In this case, there are no common factors, so there is no hole in the function. **Coordinate of Hole:** None ### f. Domain: The domain of the function is all real numbers except where the denominator is zero: \[ f(x) \in \mathbb{R} \text{ except } x = -\frac{1}{2} \] **Domain:** \( x \in \mathbb{R}, x \neq -\frac{1}{2} \) ### g. Range: The range of the function is all real numbers since the function can take any value as \( x \) approaches the vertical asymptote. **Range:** \( f(x) \in \mathbb{R} \) ### Summary: - **Vertical Asymptote:** \( x = -\frac{1}{2} \) - **End Behavior Asymptote:** No horizontal asymptotes - **\( x \)-intercepts:** \( x = -2 \) and \( x = 1 \) - **\( y \)-intercept:** \( 4 \) - **Coordinate of Hole:** None - **Domain:** \( x \in \mathbb{R}, x \neq -\frac{1}{2} \) - **Range:** \( f(x) \in \mathbb{R} \)