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
, 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
, {−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}

Answer 1

To solve the equation \( x^2 + 4x - 1 = 2x + 2 \), we can rearrange it into a standard quadratic form by bringing all terms to one side.

Starting with the equation: \[ x^2 + 4x - 1 - 2x - 2 = 0 \] This simplifies to: \[ x^2 + 2x - 3 = 0 \]

Now we can create two functions based on the left and right sides of the original equation:

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

We need to find the intersection points of these two functions, which are the values of \( x \) such that \( f(x) = g(x) \).

We already set \( f(x) = g(x) \) when we rearranged the original equation, resulting in: \[ x^2 + 2x - 3 = 0 \]

To find the roots of this quadratic equation, we can factor it: \[ (x + 3)(x - 1) = 0 \] Setting each factor to zero gives us: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

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

Putting it all together, the correct response is:

• \( f(x) = x^2 + 4x - 1 \)
• \( g(x) = 2x + 2 \)
• Solution set: \( {-3, 1} \)

So the final answer is: f(x)=x^2+4x−1, g(x)=2x+2, {−3,1}.