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?

Question

x2+4x−1=2x+2

(1 point)
Responses

f(x)=x2+4x−1, g(x)=2x+2, {−3,1}
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 1 right brace

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−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

Answer 1

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

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

This simplifies to:

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

Now we can create two functions:

1. \( f(x) = x^2 + 2x - 3 \)
2. \( g(x) = 0 \)

To find the intersection points of these two functions, we need to solve for when \( f(x) = 0 \):

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

Factoring the quadratic equation:

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

Setting each factor equal to zero gives us:

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

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

Therefore, the correct response is:

f(x)=x²+4x−1, g(x)=2x+2, {−3,1}.