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?

x2+4x−1=2x+2

(1 point)
Responses

f(x)=x2+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)=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)=x2+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−1
, g(x)=2x+2
, {−3,1}

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \), we can create two functions based on the left and right sides of the equation:

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

The intersection points of these two functions are the solutions to the equation \( f(x) = g(x) \).

To find the intersection points, we set the two functions equal to each other:

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

Rearranging the equation gives us:

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

This simplifies to:

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

Next, we can factor this quadratic equation:

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

Setting each factor to zero gives the solutions:

\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Therefore, the solution set is:

\[
\{-3, 1\}
\]

The appropriate choice from the responses provided is:

**f(x)=x²+4x−1, g(x)=2x+2, {−3, 1}**.