Question

To show how to solve the equation, create 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 +4x−1=2x+2 (1 point) Responses f(x)=x^2 +4x−1 , g(x)=2x+2 , {−1,3} f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 1 comma 3 right brace f(x)=x2+4x , g(x)=2x , {−3,1} f left parenthesis x right parenthesis equals x squared plus 4 x , g left parenthesis x right parenthesis equals 2 x , left brace negative 3 comma 1 right brace f(x)=x^2 +4x−1 , g(x)=2x+2 , {−3,−4,1,4} f left parenthesis x right parenthesis equals x squared plus 4 x minus 1 , g left parenthesis x right parenthesis equals 2 x plus 2 , left brace negative 3 comma negative 4 comma 1 comma 4 right brace f(x)=x^2 +4x−1 , g(x)=2x+2 , {−3,1}

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \), we can rewrite it to set it to zero:

\[
x^2 + 4x - 1 - 2x - 2 = 0
\]

This simplifies to:

\[
x^2 + 2x - 3 = 0
\]

Now, we can create two functions based on the original equation:

1. \( f(x) = x^2 + 4x - 1 \)
2. \( g(x) = 2x + 2 \)

To find the intersection points, we need to find the values of \( x \) where \( f(x) = g(x) \).

From the simplified equation, we can factor or use the quadratic formula:

\[
x^2 + 2x - 3 = 0
\]

Factoring gives:

\[
(x + 3)(x - 1) = 0
\]

Thus, the solutions are:

\[
x + 3 = 0 \implies x = -3
\]
\[
x - 1 = 0 \implies x = 1
\]

Therefore, the solution set is \( \{-3, 1\} \).

So the correct response is:
- \( f(x) = x^2 + 4x - 1 \), \( g(x) = 2x + 2 \), and the solution set is \( \{-3, 1\} \).