To show how to solve the equation, ereate two functions and find their intersection points. What two functions can be used to solve the following equation, and what is their solution set? \( x^{2}+4 x-1=2 x+2 \) A point) \( f(x)=x^{2}+4 x-1, g(x)=2 x+2,\{-3,-4,1,4\} \) \( f(x)-x^{2}+4 x-1, g(x)-2 x+2,\{-3,1\} \) \( f(x)-x^{2}+4 x, g(x)=2 x,\{-3,1\} \) \( f(x)-x^{2}+4 x-1, g(x)=2 x+2,\{-1,3) \)

To solve the equation \(x^{2}+4x-1=2x+2\), we can create two functions and find their intersection points. Let's define the two functions: 1. \(f(x) = x^{2} + 4x - 1\) 2. \(g(x) = 2x + 2\) Now, we need to find the intersection points of these two functions by setting them equal to each other and solving for \(x\). Setting \(f(x) = g(x)\), we have: \[x^{2} + 4x - 1 = 2x + 2\] Simplifying the equation: \[x^{2} + 2x - 3 = 0\] Now, we can solve this quadratic equation to find the intersection points. Let's solve the quadratic equation \(x^{2} + 2x - 3 = 0\). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}+2x-3=0\) - step1: Factor the expression: \(\left(x-1\right)\left(x+3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-1=0\\&x+3=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=1\\&x=-3\end{align}\) - step4: Rewrite: \(x_{1}=-3,x_{2}=1\) The solutions to the equation \(x^{2} + 2x - 3 = 0\) are \(x = -3\) and \(x = 1\). Therefore, the intersection points of the two functions \(f(x) = x^{2} + 4x - 1\) and \(g(x) = 2x + 2\) are \(x = -3\) and \(x = 1\). The correct answer is: \( f(x) = x^{2} + 4x - 1, g(x) = 2x + 2, \{-3, 1\} \)